Post on 26-Apr-2017
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: williams@eng.uah.edu
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.
5/21/2010 1
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
5/21/2010 2
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
5/21/2010 5
Modern Electromechanical DevicesModern Electromechanical Devices
5/21/2010 6
Interdisciplinary Approach to El t h i l S t E i iElectromechanical Systems Engineering
5/21/2010 7
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?
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
5/21/2010 16
the voltage, V, via integration– Obtain C=Q/V
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
Inductors and Inductance
• We now know that closed magnetic circuit carrying current I produces a magnetic field with flux
Inductors and Inductance
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Inductors and Inductance
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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.
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
Maxwell’s Time Dependent Equations:Identity MapIdentity Map
5/21/2010 38
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
Classification of Magnetic Materials (1)
• In general we use the magnetic susceptibility (or relative permeability) to classify materials in terms of their magnetic property
Classification of Magnetic Materials (1)
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Classification of Magnetic Materials (2)
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
Classification of Magnetic Materials (3)
• FerromagnetismO i t ith l ti l l ti t
Classification of Magnetic Materials (3)
– 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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Classification of Magnetic Materials (4)
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Review of Newtonian Mechanics
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
Review of Newtonian Mechanics
• 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
Review of Newtonian Mechanics
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
Review of Newtonian Mechanics
• In electromechanical motion devices, the force and torque are of great interest.• Assuming a rigid body and a constant moment of inertia one has
Review of Newtonian Mechanics
• 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.
• Derive the equations of motion using Lagrangian Dynamics
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
__
,
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
tuutduLawsKircchoffofnApplicatioThrough
CqqqR
Ldtdiq
CRq
C
a
21222
1
)(1)(_'___
1
)(
Cqq
Potential
qqdt
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
• Derive the equations of motion using 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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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!!!!!!
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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,,
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
y
N
iyyLi
xWF im
l
i
2
21)(
21
5/21/201074
Force on Magnetic Materials(E l 3 )
• 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
Figure from: M.N.O. Sadiku, Elements of Electromagnetics 4th ed. Oxford University Press, 2007.
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
Force on Magnetic Materials(E l 3 )
• 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
Force on Magnetic Materials(E l 4)
• 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
Force on Magnetic Materials(E l 5)
• 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