EE 410/510: Electromechanical SystemsElectromechanical Systems 410... · EE 410/510:...

EE 410/510: Electromechanical SystemsElectromechanical SystemsT/Th 12:45 – 2:05 PM TH N155

Instructor: J.D. Williams, Assistant Professor Electrical and Computer EngineeringU i it f Al b i H t illUniversity of Alabama in Huntsville

406 Optics Building, Huntsville, Al 35899Phone: (256) 824-2898, email: [email protected]

C t i l t d UAH A l t b itCourse material posted on UAH Angel course management website

Textbook: S.E. Lyshevski, Electromechanical Systems and Devices, CRC Press, 2008

ISBN Number: 978‐1‐4200‐6972‐3 Optional Reading:

H.D. Chai, Electromechanical Motion Devices, Prentice Hall, 1998 S.J. Chapman, Electric Machinery Fundamentals, 4th ed. McGraw Hill, 2005

S E Lyshevski Engineering and Scientific Computations using MATLAB Wiley 2003S.E. Lyshevski, Engineering and Scientific Computations using MATLAB, Wiley, 2003 A.E. Fitzgerald, C. Kingsley, S.D. Umans, Electric Machinery, 6th ed. McGraw Hill, 2003

C.W. de Silva, Mechatronics: an Integrated Approach, CRC Press, 2004

All figures taken from primary textbook unless otherwise cited.

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EE 410/510 - Electromechanical Systems:C M t i lCourse Material

• Chapter 1: Introduction to Electromechanical Systems

• Chapter 2 Analysis of Electromechanical– Torque Characteristics – 3 Phase induction motors Chapter 2. Analysis of Electromechanical

Systems – Review of Electromagnetics– Review of Classical Mechanics – Introduction to MATLAB and Simulink

– Introduction to Quadrature and Direct Variables – Arbitrary Reference Frames – Simulation of 2 and 3 Phase AC Induction Motors

using MATLAB and Simulink• Chapter 6 Synchronous Machines (advanced• Chapter 3. Introduction to Power Electronics

– Modeling and Application of Op. Amps., Power Amplifiers, and Power Converters

• Chapter 4. DC Electric Machines and Motor Devices

• Chapter 6. Synchronous Machines (advanced topic)

– Introduction – Single and Three Phase Reluctance Motors – Two and Three Phase Permanent Magnet Devices

– Geometry and Equations of Motion Governing DC Electric Motors

– Modeling and Simulation of DC Electric Motors – Permanent Magnet DC Generator

gSynchronous Motors and Stepper Motors

– MATLAB and Simulink Simulations • Chapter 7. Introduction to Control of

Electromechanical Systems and PID Control Laws– DC Electric Machines with Power Electronics

– Axial Topology of DC Electric Machines and Magnetization Currents

• Chapter 5. Induction Machines (some advanced topics)

Laws – Equations of Motion Governing the Dymamics of

Electromechanical Systems – Analog PID Control laws and application involving

Permanent Magnet DC Motor advanced topics)

– Overview 2 Phase AC Induction Motors – Equations of motion for 2 Phase AC Induction

Motors

– Digital PID Control Laws and application involving Servosystem with Permanent Magnet DC Motor

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EE 410/510 - Electromechanical Systems:C A i tCourse Assignments

Homework: Homework will be assigned throughout the semester and is due 7 days afterassignment. Assignments will be graded and returned to account for 30% of thefinal course gradefinal course grade.

Exams: Two in class exams will be given during the semester. Students will be allowedthe use of a calculator during the exam. All work will be performedindependently. Each exam will account for 25% of the student’s grade. The finalexam will be comprehensive covering major topics presented throughout thesemester and will constitute 20% of the course grade.semester and will constitute 20% of the course grade.

Final Grade: Homework Weekly 30% E 2 S t 25%Exams 2 per Semester 25% Final Comprehensive 20% 5/21/2010 3

EE 410/510: Electromechanical SystemsChapters 1 and 2

Ch t 1 I t d ti t• Chapter 1: Introduction to Electromechanical Systems

• Chapter 2: Analysis of Electromechanical Systems and Devices

• Introduction to analysis and modeling• Energy conversion and Force

Production• Overview of electromagnetics• Overview of electromagnetics• Overview of classical mechanics

(Newtonian mechanics only)• Applications of combined systems• Simulation of systems in the

MATLAB environment

All figures taken from primary textbook unless otherwise cited.5/21/2010 4

Interdisciplinary Approach to El t h i l S t E i iElectromechanical Systems Engineering

dfgdf

dfgdf

dfgdf

dfgdf

dfgdf

Ki ti EKinetic Energy

Potential Energy

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Modern Electromechanical DevicesModern Electromechanical Devices

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Interdisciplinary Approach to El t h i l S t E i iElectromechanical Systems Engineering

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Point Charge Distributions and C l b’ LCoulomb’s Law

• The force, F, between two point charges Q1 and Q2 is:– Along the line joining them– Directly proportional to the product between them– Inversely proportional to the square of the distance between them

221

4 RQQFo

– The equation above is easily calculated for a test charge, Q1, at the

Where , 2212 /10854.8 NmCo

o

k41

origin and a source charge, Q2, at a distance R away.– The solution is slightly more complicated as we move our reference

frame away from the two charge such that Q1 is referenced by the vector r1 and Q2 is referenced by the r2

5/21/2010 8

vector r1 and Q2 is referenced by the r2.

Gauss’ LawGauss Law• Gauss’ Law: The electric flux, , through any closed surface is equal to the

total charge enclosed by that surface, thus =Qenc

Integral Formencv QdvSdD

Applying the divergence theorem, we have

dvdvDSdD v

encv

vs

yielding the differential form

v

vvs

vD

This is the first of the 4 Maxwell Equations which clearly states that the volume charge density is equal to the divergence of the electric flux density

D

B

0Maxwell’s

5/21/2010 9tBE

D v

tDJH

Maxwell sEquationsIn matter

Gauss’ Law(2)Gauss Law(2)• Gauss’ law is simply an alternative statement of Coulomb’s law.• Gauss’ law provides an easy means of finding E or D for symmetrical charge

distributions• Applications:

aDD Point Charge

enc

r

r

SdDQ

ddraSd

aDD

2 sin

rrenc

senc

ddraaDQ

SdDQ

22

sin

enc rDddrDQ 222

0 0

0 0

4sin

5/21/2010 10rar

QD 24

Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.

Electric PotentialElectric Potential• We define the electric potential (or the irrotational scalar field), V, describing any

l t t ti t fi ld E th it d f th diff b t E t telectrostatic vector field, E, as the magnitude of the difference between E at two points a and b and some standard (common) reference point o.

p

ldEpV

)( negative by convention

b

a

o

a

b

o

o

ldEldEldEaVbV

p

)()(

)( ega e by co e o

• Now, the Gradient theorem states that

aao

ldEdV

Note the crucial role that independence of path plays: If E was dependent on path, then the definition of V would be

EdV

EdldV

cos

nonsense because the path would alter the value of V(p)VE

5/21/2010 11

Edl

max

The Dielectric Constant• It is important to note that up to this point, we have not committed ourselves to the

cause of the polarization, P. We dealt only with its effects. We have stated that the polarization of a dielectric ordinary results from an electric field which lines up the atomic or molecular dipoles.

• In many substances, experimental evidence shows that the polarization is proportional to the electric field, provided that E is not too strong. These substances are said to have a linear, isotropic dielectric constant

• This proportionality constant is called the electric susceptibility, e. The convention is t t t th itti it f f f th l t i tibilit t k th itto extract the permittivity of free space from the electric susceptibility to make the units dimensionless. Thus we have

• From the previous slideEP eo

ED

EPED

ro

eoo

)1( The dielectric constant (or relative permittivity) of the material, r, is the ratio of the permittivity to that of free space

• If the electric field is too strong, then it begins to strip electrons completely from molecules leading conduction effects. This is called dielectric breakdown.

• The maximum strength of the electric that a dielectric can tolerate prior to which

ED

5/21/2010 12

The maximum strength of the electric that a dielectric can tolerate prior to which breakdown occurs is called the dielectric strength.

Using Gauss’ Law With DielectricsUsing Gauss Law With DielectricsConcentric Conducting Spheres with radius a b (b>a) with a dielectric fill

Two flat conductive plates of area A filled with dielectric

aEEE s

radius a, b (b>a) with a dielectric fillarea, A, filled with dielectric

z

R

aQEo

R

4 2

-+

+-

QSdD

aEED

aEEE

enc

zsrtrott

zo

st

21z

o

s aE

21

s aE

a

b

r

Q

ldDV

b

o

AQdldDV

QDAdaDS

zo

s aE22

d

br

Q

drr

Q

a ro

11

4 2

Ad

ba4

5/21/2010 13

If capacitance, C=Q/V, then what is the value for each example?


Continuity EquationContinuity Equation • Remembering that all charge is conserved, the time rate of decrease of charge

within a given volume must be equal to the net outward flow through the surfacewithin a given volume must be equal to the net outward flow through the surface of the volume

• Thus, the current out of a closed surface is

ddQSdJI venclosed

dvt

dvJSdJ

dvtdt

QSdJI

v

v

venclosed

S

Applying Stokes Theorem

tJ

t

v

vvS

Continuity Equation

• For steady state problems, the derivative of charge with respect to time equals zero, and thus gradient of current density at the surface is zero, showing that there can be no net accumulation of charge.

5/21/2010 14

Electrical ResistivelyElectrical Resistively• Consider a conductor whose ends are maintained at a potential difference ( i.e. the

l t i fi ld ithi th d t i d fi ld i d th h th t i l )electric field within the conductor is nonzero and a field is passed through the material.)• Note that there is no static equilibrium in this system. The conductor is being fed

energy by the application of the electric field (bias potential)• As electrons move within the material to set up induction fields, they scatter and are p , y

therefore damped. This damping is quantified as the resistance, R, of the material.• For this example assume:

– a uniform cross sectional area S, and length l.The direction of the electric field E produced is the same as the direction of flow of positive– The direction of the electric field, E, produced is the same as the direction of flow of positive charges (or the same as the current, I).

lVE

ldE

1

llVR

lVE

SIJ

c

s

v

SdE

ldE

IVR

5/21/2010 15

SSIR


CapacitanceCapacitance• Capacitance is the ratio of the magnitude of charge on two separated plates

to the potential difference between them

ldE

SdEVQC

• Note that The negative sign is dropped in the definition above because we are interested in the absolute value of the voltage drop

• Capacitance is obtained by one of two methods

ldEV

– Assuming Q, and determine V in terms of Q– Assuming V, and determine Q in terms of V

• If we use method 1, take the following steps– Choose a suitable coordinate system– Let the two conducting plates carry charges +Q and –Q– Determine E using Coulomb’s or Gauss’s Law and find the magnitude of

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the voltage, V, via integration– Obtain C=Q/V


Capacitance vs ResistanceCapacitance vs. Resistance

ldEVR

Parallel PlatesS

dRdSC

,

SdEQ

SdEIR

Lab

RbLC

Sd

2

ln,2

Coaxial Cylinders

RC

ldEVQR

ba

Lab

114

2ln

Between 2 Spheres

RC

Isolated Sphere

baR

ba

C

1

4,

114

Between 2 Spheres

5/21/2010 17

Isolated Spherea

RaC

4

1,4

Summary Diagram of ElectrostaticsSummary Diagram of Electrostatics

1

dr

V v

o41

draE rv

o24

1

0

E

Eo

v

o

vV

2

VE

EV

5/21/2010 18 ldEV

EV

Figure is recopied from Griffiths, Introduction to Electrodynamics,3rd ed., Benjamin Cummings, 1999.

Biot-Savart’s Lawot Sa a t s a

• The differential magnetic field intensity, dH, produced at a point P, by the differential t l t Idl i ti l t th d t Idl d th i f th l b tcurrent element, Idl, is proportional to the product Idl and the sine of the angle between

the element and the line joining P to the element and is inversely proportional to the square of the distance, R, between P and the element

232 4sin

44ˆ

RIdl

RRlId

RalIdHd R

19


Ampere’s Circuit Law• Ampere’s law: The line integral of H around a closed path is the same as the net

current, Ienc, enclosed by the path

p

– Similar to Gauss’ law since Ampere’s law is easily used to determine H when the current distribution is symmetrical

encIldH

current distribution is symmetrical– Ampere’s law ALWAYS holds, even if the current distribution is NOT symmetrical,

however the equation is only used effectively for symmetric cases– Like Gauss and Coulomb’s Laws, Ampere’s aw is a special case of the Biot-Savart

l d b d i d di tl f itlaw and can be derived directly from it.

• Applying Stokes’s theorem provides alternative solution methods

SdHldHI

SdJI

SdHldHI

Senc

SLenc

Definition of Current provided in Chapter 5

20JH

S

Maxwell’s 3rd Eqn.

Displacement Current• Lets now examine time dependent fields from the perspective on Ampere’s Law.

p

tJ

JH

JH

v

0

0 This vector identity for the cross product is mathematicallyvalid. However, it requires that the continuity eqn. equalszero, which is not valid from an electrostatics standpoint!

D

JJH

JJHt

d

d

0

, p

Thus, lets add an additional current density term to balance the electrostatic field requirement

tDJ

tDD

ttJJ

d

vd

We can now define the displacement current density as the time derivative of the displacement vector

tDJH

Another of Maxwell’s for time varying fields

This one relates Magnetic Field Intensity to conduction and displacement current densities

5/21/2010 21

p

Magnetic Flux Density

• Magnetic Flux density, B, is the magnetic equivalent of the electric flux

g y

g y g qdensity, D. As such, one can define

mH

HB

/104 70

0

where

• Similarly, Ampere’s Law is

• And the Magnetic flux through a surface is

ldBI encˆ

0

SdHSdB

0And the Magnetic flux through a surface is

• The magnetic flux through an enclosed system is

SS

0

dvBSdB

0

B

dvBSdBSS

Definition of a solenoidal fieldand Maxwell’s 4th eqn

22

and Maxwell s 4 eqn.

Faraday’s Law (1)Faraday s Law (1)• We have introduced several methods of examining magnetic fields in terms of forces,

energy, and inductances.M ti fi ld t b di t lt f h i th h t d• Magnetic fields appear to be a direct result of charge moving through a system and demonstrate extremely similar field solutions for multipoles, and boundary condition problems.

• So is it not logical to attempt to model a magnetic field in terms of an electric one? This is the question asked by Michael Faraday and Joseph Henry in 1831 The result is Faraday’sthe question asked by Michael Faraday and Joseph Henry in 1831. The result is Faraday s Law for induced emf

• Induced electromotive force (emf) (in volts) in any closed circuit is equal to the time rate of change of magnetic flux by the circuit dd g g y

where, as before, is the flux linkage, is the magnetic flux, N is the number of turns in the inductor, and t represents a time interval. The negative sign shows that the induced voltage

dtdN

dtdVemf

, p g g gacts to oppose the flux producing it.

• The statement in blue above is known as Lenz’s Law: the induced voltage acts to oppose the flux producing it.

• Examples of emf generated electric fields: electric generators, batteries, thermocouples, fuel

5/21/2010 23

cells, photovoltaic cells, transformers.

Faraday’s Law (2)Faraday s Law (2)• To elaborate on emf, lets consider a battery circuit.

• The electrochemical action within results and in emf produced electric field, Ef

• Acuminated charges at the terminals provide an electrostatic field Ee that also exist that counteracts the emf generated potential

EEEP

ef

• The total emf generated in by a time dependent motion or magnetic field is

IRldEldEldEN

fL

fL

0

ldBuBdldEV f

• Note the following important facts

BudtBdE

ldBudt

ldEV

m

LSLemf

g p• An electrostatic field cannot maintain a steady current in a close circuit since

• An emf-produced field is nonconservative

L

e IRldE 0

5/21/2010 24

An emf produced field is nonconservative• Except in electrostatics, voltage and potential differences are usually not equivalent


Inductors and Inductance

• We now know that closed magnetic circuit carrying current I produces a magnetic field with flux


magnetic field with flux

• We define the flux linkage between a circuit with N identical turns as

SdB

• As long as the medium the flux passes through is linear (isotropic) then then flux linkage is proportional to the current I producing it and can be written as

N

Where L is a constant of proportionality called the inductance of the circuit. A circuit that contains inductance is said to be an inductor.

O t th i d t t th ti fl f th i it

LI

• One can equate the inductance to the magnetic flux of the circuit as

where L is measured in units of Henrys (H) = Wb/AI

NI

L

• The magnetic energy (in Joules) stored by the inductor is expressed as

5/21/2010 252

21 LIWm


Inductors and Inductance• Since we know that magnetic fields produce forces on nearby current elements, and that those

magnetic fields can be generated by an isolated or coupled set of current carrying circuits, then it is only reasonable that such circuits may induce fields and magnetization between them


y y g

• We can calculate the individual flux linkage between the two components as

Lik i d t i t l i d t b t th i it th t i l f i it 12

1

212S

SdB

• Likewise we can determine a mutual inductance between the circuits that is equal from circuit 12 as it is from circuit 21 as

• Individual inductances are2

121

2

1212 I

NI

M

11111

NL

22222

NL

• The total magnetic energy in the circuit is5/21/2010 26

111 II

L22

2 IIL

2112222

2111221 2

121

21 IIMILILWWWWm


Inductors and Inductance (2)• As we eluded to before, you should think of an inductor as a conductor shaped in such a way as

to store magnetic energy• Typical examples include toroids solenoids coaxial transmission lines and parallel-wire

Inductors and Inductance (2)

Typical examples include toroids, solenoids, coaxial transmission lines, and parallel wire transmission lines

• One can determine the inductance for a given geometry using the following technique– Choose a suitable coordinate system– Let the inductor carry current ILet the inductor carry current, I– Determine B from Biot-Savart’s or Amperes Law and calculate the magnetic flux– Find L as a function of the flux times the number of turns over the current carried

• Mutual inductance may be calculated by a similar approachDetermine the internal inductance L for the flux generated by the first inductor– Determine the internal inductance, Lin for the flux generated by the first inductor

– Determine the external inductance, Lext produced by the flux external of the first inductor– The sum of the internal and external inductance equals the individual inductances plus the

mutual inductance between the elements

SdB

12112 N

• For circuit theory, we can also right the inductance as which provides a very useful equation when quickly mapping out electronic circuits

1

212S

SdB2

121

2

1212 I

NI

M

CL t

RC

5/21/2010 27

CLext RC

1

extext LR

LR

Forces Due to a Magnetic FieldForces Due to a Magnetic Field• Recall that the force on a charged particle is simply F=qE

If the particle moves however then an additional force is imposed from the charge• If the particle moves however, then an additional force is imposed from the charge displacement of velocity, u, quantified by the magnetic field, B. The combined force is called the Lorentz Law:

)( BuEqF

• Recall from Newton’s Law that

)( BuEqF

dudmamBuEqF

)(

• The kinetic energy of a charged particlein an magnetic field is therefore

dtq )(

)( udmBvqF

For B, u, and a in orthogonal directions,One can deduce a coordinate system in which

)( 11

dtvdt

BvqdtBvqu

)()(

)()(

)(

dtm

BvBvqdtm

Bvqu

dtm

BvBvqdt

mBvqu

dtmBvqF

xzzxyy

zyzyxx

0)(

ˆ)(

33

212

2

1

dtm

Bvqu

uudtm

Bvqu

dtvdtm

dtm

u

mBq

Cyclotron Resonance Frequency

5/21/2010 282

21

)()(

umKE

dtm

BvBvqdt

mBvqu

mmxyyxz

z

dtldu

The location of the particle can also be found asm Frequency

dtul ii

Lorentz Force LawLorentz Force Law• Recall that the force on a charged particle is simply F=qE

If the particle moves however then an additional force is imposed from the charge• If the particle moves however, then an additional force is imposed from the charge displacement of velocity, u, quantified by the magnetic field, B. The combined force is called the Lorentz Law:

)( BuEqF

• Recall from Newton’s Law that

)( BuEqF

dudmamBuEqF

)(

• The kinetic energy of a charged particlein an electric field is therefore

dtq )(

)( udEF

ld

The location of the particle can also be found as

)(

dtqE

u

dtm

qEu

dtmEqF

yy

xx

dtul

dtldu

ii

5/21/2010 292

21 umKE

dtm

qEu

mz

z

y

Magnetic Torque and MomentMagnetic Torque and Moment

• Now that we have examined the force on a current carrying loop. Let’s examine the Torque applied to itapplied to it

• Torque, T, on the loop is the vector product of the force, F, and the moment arm, r.

IBlF

Buniformforand

___

but

IBlwT

IBlF

sin

0

IBST

yieldingSlw

sin FrT

Where we can now define a quantity m assin

0

where

FrT

l

Where we can now define a quantity m as the magnetic dipole moment with units A/m2

which is the product of the current and area of the loop in the direction normal the surface area defined by the loop

5/21/2010 30BmT

aISm n

ˆ0ˆˆ 000

FFBadzBadzIBlIdF

lzz

L


Torque and Dipole Properties f B M t

• A bar magnet or small filament loop is generally referred to as a magnetic dipoleA b ti f l th l t if ti fi ld B d di l

of a Bar Magnet

• Assume a bar magnetic of length, l, generates a uniform magnetic field, B, and a dipole moment, |m|=Qml

• Torque, T, on the loop is the vector product of the force, F, and the moment arm, r.

Thus because the bar magnet represents aThus, because the bar magnet represents a magnetic dipole moment equal in magnitude to the dipole moment of a current loop, a bar magnet can also be taken as a magnetic dipole

BlQBmT m

ISlQ

ISBlBQT

BQF

m

m

m

Therefore the field at a reasonable distance

5/21/2010 31

Therefore the field at a reasonable distance away from any bar magnet is mathematically identical to that of a dipole.


Maxwell’s Eqns for Static FieldsMaxwell s Eqns. for Static Fields

Differential Form Integral Form Remarks

Gauss’s Law

Nonexistence of the

D v

dvSdD vS

Nonexistence of the Magnetic Monopole

Conservative natureE

B

0

0 0 SdBS

0 ldE Conservative nature

of the Electric Field

Ampere’s LawJH

E

0

SdJldH

0L

ldE

5/21/2010 32

Ampere s LawJH SdJldHSL

Maxwell’s Time Dependent Equations• It was James Clark Maxwell that put all of this together and reduced electromagnetic field

theory to 4 simple equations. It was only through this clarification that the discovery of electromagnetic waves were discovered and the theory of light was developed. The equations Maxwell is credited with to completely describe any electromagnetic field• The equations Maxwell is credited with to completely describe any electromagnetic field (either statically or dynamically) are written as:

Differential Form Integral Form Remarks

Gauss’s Law

Nonexistence of theB

D v

0

dvSdD vS

Nonexistence of the Magnetic Monopole

Faraday’s LawBE

B

0 0 SdBS

Faraday s Law

D

tBE

D

SL

SdBt

ldE

5/21/2010 33

Ampere’s Circuit Law

tDJH

Sd

tDJldH

SL

Analogy Between Electric and Magnetic Fieldsa ogy et ee ect c a d ag et c e ds

• Basic Laws

Electric Magnetic

rar

QQF 211 ˆ

4

RaIdlBd r

20

4ˆ

• Force Law• Source Element

enc

dQEQF

QSdD

lIduQ

BuQF

IldHR

enc

4

• Field intensity• Flux density• Relationship Between Fields

ED

mCS

D

mVl

VE

2/

)/(

mWb

SB

mAlIH

2/

)/(

• Potentials

L

rdlV

VEED

4

RlIdA

JVH

HB

m

4

)0(,

• Flux

dtdVCI

CVQ

SdD

dILI

LI

SdBR

4

• Energy Density• Poisson’s Eqn.

34

v

E

V

EDw

dt

2

21

JA

HBw

dtLI

E

2

21

Electromagnetic Work and PowerElectromagnetic Work and Power

e

veDDEE

vWw

22

0 221

21lim

Electric energy density

T

m

vm

www

BBHHv

Ww2

2

0 221

21lim

Magnetic energy density

T

meT

dvBHEDdvwW

www

121

Electromagnetic energy

vS

dvEdvBHEDt

SdHEP2

21

Total electromagnetic power = rate of decrease in stored energy – ohmic power dissipated

5/21/2010 35

Electrostatic Boundary ConditionsElectrostatic Boundary Conditions• Electrostatic boundary conditions for E and D crossing any material interface must

match the following conditions developed using Guass’s law and conservation of thematch the following conditions developed using Guass s law and conservation of the electric field

• Two different dielectrics characterized by 1 and 2.

QSdDD

ldEE

0 S

encv QSdDD ldEE 0

21 nn DD tt DD 21

5/21/2010 362211 nn EE

tt EE 21

21


Magnetic Boundary Conditions

• Magnetic boundary conditions for B and H crossing any material interface must match the following conditions developed using Guass’s law for magnetic fields and Ampere’s circuit law

Magnetic Boundary Conditions

following conditions developed using Guass s law for magnetic fields and Ampere s circuit law

0 SdB

IldH

IldH

IldH

21 tt HH

nn BB 21

5/21/2010 372

2

1

1

tt BB

nn HH 2211


Maxwell’s Time Dependent Equations:Identity MapIdentity Map

5/21/2010 38


Classification of Magnetic Materials (1)

• In general we use the magnetic susceptibility (or relative permeability) to classify materials in terms of their magnetic property


terms of their magnetic property• A material is said to be nonmagnetic if there is no bound current density or zero

susceptibility. Otherwise it is magnetic• Magnetic materials may be grouped into three classes, diamagnetic, paramagnetic, and

ferromagneticferromagnetic• For many practice purposes, diamagnetic and paramagnetic materials exhibit little to no

magnetic susceptibility. What magnetic properties these materials do have follows a linear response over a large range of applied fields

• Ferromagnetic materials kept below the Curie temperature exhibit very large nonlinear g p p y gmagnetic susceptibility and are used for conventional magnetic device applications

5/21/2010 39


Classification of Magnetic Materials (2)• Diamagnetism

– Occurs when the magnetic fields in the material due to individual electron moments cancels each other out Thus the permanent magnetic moment of each atom is zero


cancels each other out. Thus the permanent magnetic moment of each atom is zero.– Such materials are very weakly affected by magnetic fields.– Diamagnetic materials include Copper, Bismuth, silicon, diamond, and sodium chloride

(table salt)– In general this effect is temperature independent Thus for example there is noIn general this effect is temperature independent. Thus, for example, there is no

technique for magnetizing copper– Superconductors exhibit perfect diamagnetism. The effect is so strong that magnetic

fields applied across a superconductor do not penetrate more than a few atomic layers, resulting in B=0 within the material

• Paramagnetism– Materials whose atoms exhibit a slight non-zero magnetic moment– Paramangetism is temperature dependent– Most materials (air, tungsten, potassium, monell) exhibit paramagnetic effects that provide ( , g , p , ) p g p

slight magnetization in the presence of large fields at low temperatures

5/21/2010 40


• FerromagnetismO i t ith l ti l l ti t


– Occurs in atoms with a relatively large magnetic moment– Examples: Cobalt, Iron, Nickel, various alloys based on these three– Capable of being magnetized very strongly by a magnetic field– Retain a considerable amount of their magnetization when removed from the field– Lose their ferromagnetic properties and become linear paramagnetic materials (non magnetic) when the g p p p g ( g )

temperature is raised above a critical temperature called the Curie temperature.– Their magnetization is nonlinear. Thus the constitutive relation B=0rH does not hold because r

depends directly on B and cannot be represented by a single value.

– Ferromagnetic shielding• Ferromagnetic materials can be used to “focus” and guide the flow of incident magnetic fields• By placing a ferromagnetic material completely around a device, one can shield said device from

an external field. This shielding occurs b/c the ferromagnet acts as a magnetic waveguide, that transmits the field around the shape of the structure and not within ittransmits the field around the shape of the structure and not within it.

5/21/2010 41


Classification of Magnetic Materials (4)• Ferromagnetism - B-H Curve

– The magnetization of a ferromagnet in an external applied field, H, is presented below. – As H is increased, the magnetic field, B, within the material increases significantly and then begins to

l B |H| h H


saturate to a value Bmax saturate as |H| approaches Hmax

– As the applied field, H, is removed, the ferromagnetic material retains some degree of its magnetization until the point at which the applied field H is completely reversed at which time the magnetic field inside the material saturates to the –Bmax

– The applied field is then increased again to generate the complete Hysteresis curve– Two other defining values are indicative of every B-H magnetization (Hysteresis) curve.

• When the applied field is maxed and then again reduced to a zero value. The magnetic field within the material remains at some positive value Br referred to as the permanent flux density.

• The value upon which B become zero under an applied H value is called the coercive field intensity, Hcy c

• Materials with small coercive field intensity values are said to be soft magnetic materials and do not retain significant magnetization upon the removal of the field

• Hard magnets (permanent magnets) have very largecoercive field intensity values

5/21/2010 42


Magnetic Properties of Common MaterialsMagnetic Properties of Common Materials

5/21/2010 43

Magnetic Properties of Common MaterialsMagnetic Properties of Common Materials

5/21/2010 44

Review of Newtonian MechanicsReview of Newtonian Mechanics

• Let’s recall the study of classical Newtonian mechanics in which a Distance ˆˆˆ

yx

ayayaxrNewtonian mechanics in which a mass is moved from one location to another in a Cartesian coordinate system

• The mass is pushed with a force, F,

Distance

dzdtdydtdx

zyx

dtd

dtrd

vvv

v

zyayayaxr

z

y

x

zyx

Velocity

p , ,equal to its mass times its acceleration, a.

• The work done upon the mass is the line path independent line integral of F l th l d t L Potential2

2

2

2

2

2

2

2

2

zddt

yddt

xd

zyx

dtd

dtrd

aaa

a

dtdz

z

y

x

z

Acceleration

F along the closed contour L• And we can define the system by the

amount of energy required to move the mass at a velocity, v, plus the potential energy of the mass to

Potential Gradient

)()(

)(),( 2

2

2

rdWldrtFW

rdtpd

dtvmd

dtrdmamrtF

dtzdz

Total Force

Work Donepotential energy of the mass to move without any external allied forces by a distance r. )()(),(

,

)(),(

2

d

dtrd

dtdrr

dtrdrtF

dtdWlikewise

rdl

ldrtFWL

Work Done

5/21/2010 45

0)()()(

),(0),(

2

2

2

2

rrrdt

rdm

rtFdt

rdmrtFam

Energy of the system: Kinetic and Potential

Examples of Mechanical Force and WorkKinetic EnergyPotential Energy

Examples of Mechanical Force and Work

• Example 1C id iti i t bl t t d b t– Consider a positioning table actuated by a motor

– Let us find out how much work is done to accelerate a 20g payload with mass = 20g from v0=0 to vf= 1 m/s

JmvmvW 01001020011 22

• Example 2– Consider a mass, m, slid across a flat surface in a Cartesian coordinate system

JmvmvW f 01.001020.022 0

– The force , F, is applied and motion occurs in the x direction– Find the equations of motion (neglecting coulomb and static friction), however include

viscous friction

vBxdBF

112

2

vBFFFxdvafra

2

32

0

)42cos(4

ydFFF

tdtxdetxF

vBdt

BF

ta

vvfr

___2_

2

vdtdx

ODEslinearorderfirstyieldsmmdt vafra

5/21/2010 462

2

2 0

dtxdmmaFFF

dtymmaFFF

xfrax

ygNy

0,)42cos(41 32 tvBvtetxmdt

dvv

t

Review of Newtonian Rotational Mechanics

• One can show for rotational devices that the torque, T, generated on an object with a fixed rotation length |R| is solved in a very similar manner to a linear displacement

Review of Newtonian Rotational Mechanics

fixed rotation length |R| is solved in a very similar manner to a linear displacement and rotation

dtdJ

dtdJJtT ),( 2

2

Sum of all Torques in 3-D

dtRd

vmRpRLmomentumAngulardtdt

m

0

,_

Recall that the rotation length is fixed

JLdtdJ

dtLd

FRdtpdRpR

dtd

dtLdT

mm

m

For a 1-D system, the sum of all of the moments is:

where J is the moment of inertia (kg*m2) and is the angular acceleration of the body (rad/s2)

JM

is the angular acceleration of the body (rad/s2)

5/21/2010 47

Rotational Mechanics Example

• A motor has the equivalent moment of inertia, J=0.5 kg*m2. When the motor l t th l l it f th t i 10t3 t 0

Rotational Mechanics Example

accelerates, the angular velocity of the rotor is =10t3, t0.• Find the angular moment and the torque as a function of time.

NdkJL 332 5/1050 mNt

dtdLT

smNtsradtmkgJL

m

m

2

332

15

5/105.0

5/21/2010 48

Review of Newtonian Mechanics

• Consider the translational motion of a body which is attached to an ideal spring that t f hi h b H k ’ L N l ti f i ti bt i th


exerts a force which obeys Hooke’s Law. Neglecting friction, one obtains the following expression for energy

22

21

21 xkmvE sT

Total energy in the system

2

2

2121

22

xk

mv

s

ks is the spring constant

Kinetic Energy

Potential Energy

For rotational motion and a torsional spring

2

22

121

21

J

kJE sT

2

212

k

J

s

For translational and rotational motion and a torsional spring

22

21 Jmv

5/21/2010 49


• The moment of inertia J depends on how much mass is distributed with respect to the axis. Thus note that J is different for different axes of rotation and will have to be


recalculated as one changes coordinate systems or rotational direction• If the body has a uniform density, J can be calculated for regularly shaped bodies

using their dimensions. For example, a rigid cylinder of mass m, radius R, and length l has the following horizontal and vertical moments of inertia Jlength, l has the following horizontal and vertical moments of inertia, J

22

2

121

41

21

mlmRJ

mRJ

vertical

horizontal

• The radius of gyration can be found for irregularly shaped objects, and the moment of inertia can be easily obtained. (see in class handout)

124vertical

5/21/2010 50


• In electromechanical motion devices, the force and torque are of great interest.• Assuming a rigid body and a constant moment of inertia one has


• Assuming a rigid body and a constant moment of inertia, one has

• The total work done is given by

dJddtdJd

dtdJdJdT

g y

20

2

21

00

jjdJdTW f

ff

Kinetic Energy

• Furthermore, power is

• This is the analog of applied for translational motion

T

dtdT

dtdWP

vFP • This is the analog of applied for translational motion

• Example:– Assume the rated power and angular velocity of a motor are 1W and 1000 rad/s. The

rated electromagnetic torque is found to be

vFP

5/21/2010 51mN

sradWPT

r

001.0/1000

1

Friction in Motion Devices

• Because all or the devices discussed in this course lead to mechanical motion, a discussion of friction is essential

• Friction is a highly complex nonlinear phenomenon that is typically simplified to a series of equations that adequately map losses in the performance characteristics of the system

• For our purposes we will simplify “friction” into one of three different descriptions– Coulomb friction is a retarding force (or toque) that changes its sign with the reversal of g ( q ) g g

the direction of motion. The equations of coulomb friction are

where kfc and kTc are the Coulomb friction coefficients

dtdsignkT TcCoulomb

dtdxsignkF FcCoulomb

fc Tc

– Viscous friction is a retarding force (or torque) that is a linear (or nonlinear) function of displacement

1

12

n

n

mnmviscous dtdB

dtdBT

1

12

n

n

vnvviscous dtdxB

dtdxBF

where Bv and Bm are the Coulomb friction coefficients– Static friction exist only when the body is stationary and vanishes as motion begins

0

dtdststatic TT 0

dtdxvststatic FF

5/21/2010 52

Friction in Motion DevicesFriction in Motion Devices

• For several of the problems presented within this text, the following equation for friction will be p p , g qapplied to provide adequate mapping of physical systems with frictional memory, presliding conditions, etc is.

signvkekkF frvk

frfrfr 321

signkekkT frk

frfrfr 321

5/21/2010 53

Simple Pendulum

• A point mass is suspended by a massless unstretchable string of length l.• Derive the equations of motion

Simple Pendulum

• Derive the equations of motion

d

2

sin1

mlJ

TmglJdt

ddt

a

i Tgddtd

a

Eqns. of Motion: 2 linear ODEs

torqueappliedTmgF

sin

2sinmll

gdt

a

a

a

d

TmgldtdJJT

TmglM

1

sin

sin

2

2

2

5/21/2010 54

aTmglJdt

d sin1

2

2

Lagrangian Dynamics

• Lagrangian dynamics is an energy based technique using generalized coordinates to develop easily obtain equations of motion for even the most complicated of systems

Lagrangian Dynamics

p y q p y• It works for all energy based systems and can therefore be used readily in multi-domain

problems such as electromechanical problems that include electronic circuits, magnetic flux, torque, and even hydraulic components to the system

• We start by defining the kinetic, dissipative, an potential energy terms in the system

Kinetic Energy

nn

nn

dtdq

dtdqqqtD

dtdq

dtdqqqt

,...,,,...,,

,...,,,...,,

11

11

Dissipative Energy (non-conserative)

• Where qi are general coordinates best matched to the geometry of the system, and qdoti is the derivative of qi with respect to time (the momentum component). Q is a generalized

nqqtdtdt

,...,, 1

Potential Energy

qi p ( p ) gforce term used to meet the coordinate used.

• The Lagrange equation of motion is then defined as

iq Qdtdqq i

i

g g q

5/21/2010 55Q

qqD

qqdtd

iiii

Simple Pendulum Revisited

• A point mass is suspended by a massless unstretchable string of length l.• Derive the equations of motion using Lagrangian Dynamics

Simple Pendulum Revisited

lm

2

0

21

• Derive the equations of motion using Lagrangian DynamicsKinetic Energy

qscoordinatedgeneralize

i

:_

mgl

mgl

ml

2

sin

)cos1(Potential Energy

dtdqi

d

mgldtdlml

dtdml

dtdL

g

2 sin20

Lagrangian 0 unstretchable string

mgldtdmlT

mgldtdml

a

2

2

sin

0sin

Eqns of Motion: 2 linear ODEs

Lagrange Eqn. of Motion

Homework:Develop equations of motion for a pendulum with a point mass, m, suspended by a massless spring of nominal length, l, and spring constant k Which method is easier

dtd

mlT

lg

dtd a

2sin5/21/2010 56

Eqns. of Motion: 2 linear ODEs spring constant, ks. Which method is easier, Newtonian or Lagrangian Dynamics?

Simple Double Pendulum (1)http://scienceworld.wolfram.com/physics/DoublePendulum.html

• A point mass, m1, is suspended by a massless unstretchable string of length, l1, which in tern suspends a point mass m suspended by another massless unstretchable string of

Simple Double Pendulum (1)

111

sincos

lylx

tern suspends a point mass m2, suspended by another massless unstretchable string of length, l2.

• Derive the equations of motion using Lagrangian Dynamics

Potential energy is

22112

22112

111

sinsincoscos

sin

llyllx

ly

Which we convert to

Ki ti i th

sin

coscos

11211

22211212211

glmm

glmglmmgxmgxm

22

2221221212

21

2121

22

222

21

211

21cos

21

21

21

lmllmlmm

vvmvvm yxyx

Kinetic energy is then sin 2222

L

glm

Lagrangian is

12212122

12212121

sin

sin

22

llm

llm0

111

d

dtd

Equations of motion are

1212122222

2

12221212121

1

cos

cos

llmlm

llmlmm

57

0222

dtd

5/21/2010

Simple Double Pendulum (2)http://scienceworld.wolfram.com/physics/DoublePendulum.html

• A point mass, m1, is suspended by a massless unstretchable string of length, l1, which in tern suspends a point mass m suspended by another massless unstretchable string of

Simple Double Pendulum (2)

tern suspends a point mass m2, suspended by another massless unstretchable string of length, l2.


d

Equations of motion are

1121122121212222121222121

2121

111

0

sinsinsincos0

0

d

glmmllmllmllmlmm

dtd

222122121212212121212122

222

222

sinsinsincos210

0

glmllmllmllmlm

dt

If torques are applied to drive the system then, then the equations become2dddtd

q pp y q

222122121212

212121212122

2222

1121122121212222121222121

21211

sinsinsincos

sinsinsinsincos

glmllmllmllmlmT

glmmllmllmllmlmmT

2dtdt

58

21221112211121122222

12112212212222212222112111

sinsinsincos

sinsinsincos

glllllmT

gmmlmlmlmlmmlT

Circuit Network using Lagrangian Dynamics• Consider a 2-mesh electric circuit shown here• Derive the equations of motion using Lagrangian Dynamics

Circuit Network using Lagrangian Dynamics

iqiqsiq

siq

currentscoordinatedgeneralize

,

,

:_

2211

22

21

222

22112

211 2

121

21 qLqqLqL

Kinetic

QtuVoltageApplied a )(:_

21

21 2

22211 qRqRD

eDissipativ

Homework:1. Program this into MATLAB and simulate

the response. Compare to the result

21211211

21

0,0

222

qLqLLq

qq

__

,

22

222

111

Dd

MotionofEquations

qRqDqR

qD

presented on page 61 of your textbook

2. Derive the same set of equations for this circuit using Kirchhoff’s Law.

22

21

21221122

1

11 qqPotential

qLLqLq

q

022222

11111

Qqq

Dqqdt

d

Qqq

Dqqdt

d

2

2

21

1

1

2

2

1

1

,

21

21

Cq

qCq

q

Cq

Cq

595/21/2010

0

)(

2

2222122112

1

1112121121

CqqRqLLqL

tuCqqRqLqLL a

Simulink PerformanceSimulink Performance

5/21/2010 60

Another Example of a Circuit Network using L i D i

• Consider a 2-mesh electric circuit shown here• Derive the equations of motion using Lagrangian Dynamics

Lagrangian Dynamics

iqiqsiq

siq

currentscoordinatedgeneralize

,

,

:_

2211

22

11

qL

Kinetic

22

1

QtuVoltageApplied a )(:_

11 222

211 qRqRD

eDissipativ

tuC

qqR

iq

DEsOrderSecondandFirst

a21

1 )(1____

qLddqL

qqq

qL

22

121

2

,

0,0,0

2

__

,

22

222

111

2211

Dd

MotionofEquations

qRqDqR

qD

qq

tuutduLawsKircchoffofnApplicatioThrough

CqqqR

Ldtdiq

CRq

C

a

21222

1

)(1)(_'___

1

)(

Cqq

Potential

qqdt

qq

221

22

22

21

,

022222

11111

Qqq

Dqqdt

d

Qqq

Dqqdt

d

qiqiwhere

iRuLdt

diR

tuiRu

Cdttdu

La

LLCL

aL

Ca

21,:

1

)(1)(

Cqq

qCqq

q21

2

21

1

,

615/21/2010 0

)(

21222

2111

CqqqRqL

tuC

qqqR a

Cqqu

iidt

duC

c

Lac

21

Electromechanical Actuator using L i D i

• Consider the combined electronic and mechanical systemD i th ti f ti i L i D i

Lagrangian Dynamics


rrs qsiq

siq

scoordinatedgeneralize

321 ,,

:_

eDissipativ

Kinetic

mechelec

Lrs

rrs

TQtuQtuQqiqiq

321

321

),(),(,,

23

22

21 2

121

21

DDD

qBqRqRD

DDDeDissipativ

mrs

mechelec

__

d

MotionofEquations

222

3

23

2221

21

111coscos

21

21

21

JLLL

qLLLL

qJqLqqLqL

mutualmutualsrsr

rsrs

mechelec

23

33

222

111

21

,,

qk

Potential

qBqDqR

qDqR

qD

smechelec

m

22222

11111

Qqq

Dqqdt

d

Qqq

Dqqdt

d

23111

23

22213

21

cos0

cos,0

22cos

2

LL

qqLqLqq

qJqLqqqLqL

mutuals

rmutuals

i)(__

RLLLtMotionofEquations

3321

,0,0

2

qkqqq s

33333

Qqq

Dqqdt

d

33

2133

21322

,sin

cos,0

qJq

qqqLq

qLqqLqq

mutual

rmutual

625/21/2010

332133

22313213

11323231

sinsincos)(sincos)(

qkqBqqqLqJTqRqqqLqLqqLtuqRqqqLqqLqLtu

smmutualL

mutualrmutualr

mutualmutualss

Electromechanical Actuator using L i D i (2)

• Let’s convert our Lagrangian equations into a set of linear ODE’s using time dependent differentials of current angular velocity and angle

Lagrangian Dynamics (2)

scoordinatedgeneralize :_time dependent differentials of current, angular velocity, and angle

rrs

rrs

qiqiq

qsiq

siq

321

321

,,

,,

rmutualrmutualr

smutualmutualss

qRqqqLqLqqLtuqRqqqLqqLqLtu

MotionofEquations

sincos)(sincos)(

__

2313213

1323231

Lrs TQtuQtuQ 321 ),(),(

rr

smmutualL

dtd

thatrecallqkqBqqqLqJT

:_

sin 332133

Homework: Show the conversion between these two forms

rrmutualsrrrrmutualrrrmutualrrsrmutualssss

LLL

uLuLiLLiLRiLiLR

dtdi

followingtheYieldsdt

coscoscos2sin21

__

22

2

rmutualrs

rssrmutualrrrmutualrsrrsrmutualsrsmutualsr

rmutualrs

LLL

uLuLiLiLRiLLiLR

dtdi

LLLdt

cos

cos2sin21sincos

cos

22

2

22

635/21/2010

rr

Lrsrmrsrmutualr

dtd

TkBiiLJdt

d

sin1

Equations of Motion for an Elastic Beam

• Consider the beam supported at one end

Equations of Motion for an Elastic Beam

pp• Derive the equations of motion using Lagrangian Dynamics

lx

lxxy 3

3

2

2

321)(

Kinetic

tqlx

lxxty

ll

1 321

3

3

2

2

33111

)(321),(

2

Potential

qlAqq

qlAdxqlx

lxAdmyq

11

21

0

23

3

2

21

0

28066,0

280333

21

21

21)(

66

__

11111

EI

Qqq

Dqqdt

d

MotionofEquations

qlEIq

qlEI

lxd

lx

lqEIdx

xyEIq

Potential

3

23

1

0

2

2

21

02

2

3)(231

23

221)(

),(7.12

328066),(

4

3

xtFlA

EIdtqdq

qlEIqlAxtF

q

q

645/21/2010

lq 31 lAdt

Rotational Electrostatic Motor ExampleRotational Electrostatic Motor ExampleWLAC

g

WLVg

WFCVW

gg

eexe

2

22

2,

2

widththealongntmisalignmeforx

WLgV

xWF e

2

2

tan

____2

xW

gLV

xWF

capacitortheof

e

2

2

tan 2

__ Torsional Ratcheting Actuator A high torque

rotary electrostatic actuator http://mems.sandia.gov/about/electro-gmechanical.html

65

Rotational Electrostatic Motor ExampleRotational Electrostatic Motor Example

2

____tan_

rE

torqueforneedediscecapacilcylindrica

lr

1

2

2

ln2

2

1

rL

VQC

rrdrEVVV l

r

rrba

2

1

2

ln

2

ln

rVLC

rrV

l rr

rr

NC

1

2ln

2

1r

r

re V

rr

NVCT

r

2

1

2

2

1

ln21

LmLmer TBVN

JTBT

Jdtd

equationsmechanicalTorsional

211

__

66r

r

dtd

rrJJdt

1

2ln

Magnetic Circuits

• The following relations allow one to solve magnetic field problems in a manner similar to that of electronic circuits

Magnetic Circuits

electronic circuits• It provides a clear means of designing transformers, motors, generators, and relays using a

lumped circuit model• The analogy between electronic and magnetic circuits is provided below

5/21/2010 67

Magnetic Circuits

• To develop our model we define a magnetomotive (mmf) force that is equivalent to voltage for electronic circuits

Magnetic Circuits

electronic circuits

• We then define a reluctance which is equivalent to magnetic resistance

ldHNI

l

• And show that the magnetic flux is equivalent to current using the following Ohmic style relationSl

5/21/2010 68


Donut Shaped Toroid

• A steel toroidal core with permeability,=r0 has a mean radius, 0, and a circular cross section of diameter 2a Calculate the current required to generate a flux in the core

Donut Shaped Toroid

of diameter 2a. Calculate the current required to generate a flux, , in the core

Method 1

0

2NI

lNIB r

20

2

0

0

0

22

2

NI

aNIBdS

l

r

M th d 2

20 aN r

2

02

0

002

0

0

22222 NIa

aNI

aSlNI

r

r

r

Method 2

5/21/2010 69

20

02

0

0 22aNaNN

Irr


Toroid with Rectangular Cross-section

• A steel toroidal core with permeability,=r0 has a mean radius 0, and a rectangular cross section 2a *b Calculate the current required to generate a flux in the core

Toroid with Rectangular Cross section

section, 2a b. Calculate the current required to generate a flux, , in the core

aNIbdNIbbdNIBdS

NIlNIB

aa

r

0000

0

0

l

200

aabN

INL

aaNIbdNIbbdNIBdS

r

r

a

r

a

r

0

02

0

0

0000

ln2

ln222

00

5/21/2010 70

aaIbNLIW r

m0

02

02 ln42

1


Force on Magnetic Materials(E l 2)

• For the magnetic circuit shown below, with magnetic flux density of 1.5 Wb/m2 and a relative permeability of 50.Fi d th i di id l l t d d t i th t t l t i d t t ti l

(Example 2)

• Find the individual reluctances and determine the total current required to generate a particular flux value. All branches have a cross sectional area of 0.001 cm2

143

50

1 r

path

1031030

563516

123

84

3

2

a

l

pathandpath

path

20109.0

1010501090

20103

1010501030

84

8

40

4

3

400

21

r S

l

||

20105

10101010

21

2121

8

40

4

0

a Sl

ASBI

SBNI

Ta

a

Ta

1644104.710105.1 88

5/21/2010 7120104.7||

8

213

aT

AN

I Ta 16.4420400


Force on Magnetic Materials

• Because these are often mechanical system, it is extremely useful to be able to determine the forces generated by these circuits and those required to move components from one location to

Force on Magnetic Materials

forces generated by these circuits and those required to move components from one location to another.

• For ease of use, we will ignore fringe fields in our calculations• Also, by using ferromagnetic materials and applying simple magnetic boundary conditions, we

path very strong fields into specific geometric shapes This allows one to focus the force to apath very strong fields into specific geometric shapes. This allows one to focus the force to a specific location and move an object in a well defined manner

• Lets examine the force required to pull a magnetic bar vertically up to an electromagnetic yoke.

B 1 2

T

Tm

FFSBF

SdlBdlFdW

22

212

2/12/10

2

0

mwBHBA

Fp

SBF

21

2

22

2/12/1

0

2

2/1

Force on a single gap

5/21/2010 72

A 22 0

Note: This diagram shows a yoke pulling a bar magnet (keeper) at two gap locations

pressure = energy density!!!!!!


Force on Magnetic Materials(E l 3 )

• A U-shaped electromagnetic is designed to lift a 400 kg mass (which includes the mass of the keeper) The iron yoke with relative permeability of 3000 has a cross section of 40 cm2 and a

(Example 3a)keeper). The iron yoke with relative permeability of 3000 has a cross section of 40 cm2 and a mean length of 50 cm. Each of the air gaps are 0.1mm long. Neglecting the reluctance of the keeper, calculate the number of turns in the coil when the excitation current is 1A.

21

21

ydFdvBdvHWm

22

22

22

0

20

_,

mgSBF

FdySdyB

dWdW

a

gapairmm

v Lv

48

10600403000

50.02

/11.1

6

20

0

Sl

NI

mWbS

mgB

y

gy

a

656

48105

004.00001.02

48004.030006

00

00

NINI

Sl

S

gy

g

gy

gg

g

ry


5/21/2010 73 16215

110001.011.1

0

0

N

lBlHNI aaaa

gy

gg

gygy

Force on Magnetic Materials(E l 3b)

ydFdvBdvHW 11 • Yoke pulling a keeper from both ends

(Example 3b)

SBWF

FdydyS

SdyB

dWdW

ydFdvBdvHW

am

gapairmm

v Lvm

200

_,

2

222

22

tLiW

SmgB

mgx

F

a

aml

2

0

0

)(1

22

NItiti

NL

tLiW

i

aa

am

)()(

)(2

NyL

Sty

Sl

Sl

i

grk

kk

ry

yy

2

000

)(

)(2,,


y

N

iyyLi

xWF im

l

i

2

21)(

21

5/21/201074


• Yoke pulling a keeper by gravitational force

(Example 3c)

SmgB

mgSBx

WF

a

aml

0

0

2

22

titi

NL

tLiW

S

aa

am

a

2

)()(

)(21

llSNNtxL

Stx

Sl

Sl

NI

rkry

grk

kk

ry

yy

i

022

000

)(2))((

)(2,,

vdtdx

mgSBltxl

tiSNx

titxLx

WF

ltxl

a

kryrkryyrk

arkryaml

kryrkryyrki

0

2

2

22220

22

22

)(2)()())((

21

)(2))((

mg

ltxltiSN

mdtdv

dt

kryrkryyrk

arkry2

22220

2

)(2)(1

5/21/201075


Force on Magnetic Materials(E l 3d)

• Yoke pulling a spring loaded keeper from both ends

(Example 3d)

mgB

xkxkxkSBx

WF sssam

l

0

210

2

22

titiNL

tLiW

SB

aa

am

a

2

)()(

)(21

SNN

Stx

Sl

Sl

NI

k

grk

kk

ry

yy

i

aa

022

000

)(2,,

)()(

dx

xkxkxkSBltxl

tiSNx

titxLx

WF

ltxlSNNtxL

sssa

kryrkryyrk

arkryaml

kryrkryyrk

rkry

i

210

2

2

22220

22

0

22

)(2)()())((

21

)(2))((

5/21/201076

xkxkxkltxl

tiSNmdt

dv

vdtdx

ssskryrkryyrk

arkry212

22220

2

)(2)(1


• The system is driven by voltage control and not steady state current

(Example 3e)

d

tixLdtRid

dtRitu

xkxkxkSBltxl

tiSNx

titxLx

WF

aaaa

sssa

kryrkryyrk

arkryaml

210

2

2

22220

22

)(*)()()()(

22

)(2)()())((

21

tutviltxl

tiSNtRi

xLdttdi

dtdx

dtxdLi

dttditRitu

dtdt

aakryrkryyrk

arkrya

a

aa

aa

aaa

2

22220

2

)()()(2

)(2)(

)(1)(

)()()()(

)()()(

tutviltxl

tiSNtRi

xLdttdi

isequationsofsetcompletetheThus

aakryrkryyrk

arkrya

a

yyy

2

22220

2

)()()(2

)(2)(

)(1)(

:______

v

dtdx

xkxkxkltxl

tiSNmdt

dvsss

kryrkryyrk

arkry

212

22220

2

)(2)(1

5/21/201077


• Solenoid with stationary member anda movable plunger with drag.

(Example 4)g g

xkxkFforcerestoringspring

tFxkxkdtdxBtF

txm

sss

Lssve

1

12

2

:__

)()((1)

xxLi

xxiWxiF

ixLxiW

systemmagneticlinearFor

ce

c

2

2

)(21),(),(

)(21),(

___

dxSlSSSN

xxL

dxSlSSSNNxL

ryyg

gyr

ryyg

gyr

i

2

220

2

022

2)(

2)(

22

2__),2_(_)1_(

0222

02 u

SSNdxSlS

ivdxSlS

Si

SSNdxSlSR

dtdi

obtainsoneandCombining

gyr

ryyg

ryyg

yr

gyr

ryyg

dxxdLidixLRiu

ixLdt

dRiu

lawsKirchhoffApplying

)()(

)(,

_'_ )(

221 12

2

220

2

vdtdx

mtFv

mB

mxkxki

dxSlSSSN

dtdv Lvss

ryyg

gyr

Apply the sliding friction term described previously

5/21/201078

uivdxSlS

SSNRi

xLdtdi

dtdxi

dtxLRiu

ryyg

gyr2

220

2

2)(1

)(

(2)

)(321 vsignvkekkB frkv

frfrv

Apply the sliding friction term described previously


• Induced emf from Faraday’s law may also provide sufficient analysis to complete a preliminary design

(Example 5)design

Th t t l fl li k i l t i hi b d

rr

r

remf

termrtransformetheisdt

dwhere

dd

dtd

dtd

dd

dtd

dtdV

_____

• The total flux linkage in many electric machines can be expressed as

• For radial topology machines, one uses

ps

iN

N 41

where Ns is the number of turns, and p is the flux per pole

where i= phase current, Rinst is the inner sector radius, L is the inductance, P is the number of poles, and ge is the equivalent gap (air gap + radial thickness of the permanent magnet)

inste

sp LR

gPiN 2

• Denoting the number of turns as Ns, one has

rrssag

rs

DLNiPBT

PP

iNmmf

21

cos

Dr = outer side rotor diameterLr = axial rotor length

5/21/201079

re

sag

g

PPgiNB cos

2

2

Field in the air gap

r g

Solving 2nd order ODEsSolving 2 order ODEs

• Lets examine a few second order ODEs commonly solved2 dxxd dudiid 12

• Notice that all of these are remarkably similar resulting in similar solutions

)(

)(

2

2

2

tTkdtdB

dtdJ

tFxkdtdxB

dtxdm

asm

asv

dtdiu

Ldtdu

RdtudC

dtdui

CdtdiR

dtidL

a

a

11

1

2

2

2

• One can therefore infer that if the right choice of multiplicative constants were used, that each of these could easily be converted into another. Thus the use of electronic control to optimize translational and rotational mechanical systems as well as magnetic devices

• So in a generalized form, one can solve this ODE using the following method

02:_

)(2

22

02

2

eqnsticcharacteri

tfxdtdx

dtxd

22

:_

:_'__

rootssticcharacteridtdsequationsLaplacefromobtained

02 2120

2 ssssss

20

2

:

yielding

20

22,1 s

ftsts cbeaetx )( 21

5/21/2010 80

20

22,1

20

2

2120

2

js

ss f

t

ft

ctbtaetx

cebatx

220

220 sincos)(

)(

2nd Order ODE Example (1)2 Order ODE Example (1)

• Lets examine a series RLC circuit:dt

duiCdt

diRdt

idL a1

2

2

RR

LCs

LRs

2

2

1

01

dtCdtdt

LCLRFor

LCLLs

2

2,1

12

22

LCL

Rctbtaetx t

0

220

220

1,2

sincos)(

LCRC 0

1,2

1

Other eqns. such as a parallel RLC and a mechanical spring motion result in

5/21/2010 81mk

mkB s

s

v 0,2

2nd order ODE example (2)2 order ODE example (2)

Matlab CodeMatlab Code

5/21/2010 82

Solving Differential Equations in MatlabSolving Differential Equations in Matlab

for

adding

5/21/201083

Matlab Code

Reexamining a simple series RLC circuitReexamining a simple series RLC circuit

For constant voltage and given RLC constants

Now add initial conditions, and time for the solution[V, I] = dsolve ('DV=1/C', 'DI=(-V-R*I+Va)/L', 'V(0)=20, I(0)=-10')t=0:0.001:1;

The solution for both the voltage and current in the circuit is now solved using the Matlab ODE 45 solver as

5/21/2010 84Plot the function in Matlab using the following code

Example with Three Equations of MotionExample with Three Equations of Motion

• Equations of motion

• Initial Conditions

• Now lets examine options to code this solution

( Open Matlab )( Open Matlab )

5/21/2010 85

Van der Pol’s EqnVan der Pol s Eqn

• Nonlinear differential equation l d t di t h t h thcommonly used to predict heart rhythm,

and tunneling diodes

• Solved using coupled differential equations

• Now lets examine some code and then solve with SIMULINK

5/21/2010 86

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