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PowerPoint Slides
to accompany
Electric Machinery
Sixth Edition
A.E. Fitzgerald
Charles Kingsley, Jr.
Stephen D. Umans
Chapter 3
Electromechanical-EnergyConversion Principles
3-0
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3.1 FORCES AND TORQUES IN MAGNETIC FIELD SYSTEMS
Lorentz Force Law:
F  q (E  v  B)
For many charged particle
Fv   (E  v  B)
3-1
N/m3
(

coulombs/m3)
Current density
Jv
A/m2
Fv  J  B
Current
IJA
A
F  I  B N/m
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Example 3.1: A nonmagnetic motor containing a single-turn coil is
placed in a uniform magnetic field of magnitude B0, as shown
in Fig. 3.2. The coil sides are at radius R and the wire carries
current I as indicated. Find the θ-directed torque as a function
of rotor position α when I=10 A, B0=0.02 T and R=0.05 m.
Assume that the rotor is of length l=0.3 m.
3-2
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Very few problems can be solved using Lorentz force, where
current-carrying elements and simple structures exist.
Most electromechanical-energy-conversion devices contain
magnetic material and forces can not be calculated from Lorentz
force.
Thus, We will use ENERGY METHOD based on conservation of
energy.
3-3
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Electrical terminals: e and i
Mechanical terminals: ffld and x
Losses separated from energy storage mechanism
Interaction through magnetic stored energy
3-4
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Time rate of change of Wfld (field energy) equals to the difference
of input electrical power and output mechanical power for
lossless systems.
d W fld
dt
 e i  f fld
dx
dt
or
d Wfld  i d  f fld dx
Force can be solved as a function of flux linkage λ and position x.
3-5
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3.2 ENERGY BALANCE
Energy neither created nor destroyed, it only changes the form.
Energy balance equation is written for motor action below
Energy input Mechanical  Increasein   Energy 
 from electric   energy    energy stored   converted

 
 
 

 sources   output  in magneticfield  t o heat 
For lossless magnetic-energy-storage system
d Welec  d Wmech  d W fld
d Welec : Differential electricalenergy input
d Wmech : Differential mechanicalenergy output
d W fld : Differential change in magneticstoredenergy
3-6
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3.3 ENERGY IN SINGLY-EXCITED MAGNETIC FIELD SYSTEMS
Schematic of an electromagnetic relay. Figure 3.4
3-7
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The magnetic circuit can be described by an inductance which is a
function of the geometry and permeability of the magnetic material.
When air-gap exist in most cases Rgap>>Rcore and energy storage
occurs in the gap.
Magnetic nonlinearity and core losses neglected in practical
devices.
Flux linkage and current linearly related.
Energy equation
  L( x ) i
d Wfld  i d  f fld dx
Wfld uniquely specified by the value of λ and x. Thus, λ and x are
called STATE VARIABLES.
3-8
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Magnetic stored energy Wfld uniquely determined by λ and x
regardless of how they are brought to their final values.
W fld 
 d W fld   d W fld
path2 a
path2b
0
W fld (0 , x0 )   i( , x0 ) d
0
OR magnetic stored energy:
W fld
B

    H dB  dV
V0

Integration paths for Wfld. Figure 3.5
3-9
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Example 3.2:The relay shown on the figure is made of infinitelypermeable magnetic material with a movable plunger, also of
infinitely-permeable material. The height of the plunger is much
greater than the air-gap length (h>>g). Calculate the magnetic
stored energy Wfld as a function of plunger position (0<x<d) for
N=1000 turns, g=2 mm, d=0.15 m, l=0.1 m, and i=10 A.
3-10
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3.4 DETERMINATION OF MAGNETIC FORCE AND TORQUE
FROM ENERGY
Consider any state function F(x1, x2), the total differential of F
with respect to the two variables x1 and x2
F
d F ( x1 , x2 ) 
 x1
x2
F
d x1 
 x2
d x2
x1
Similarly, for energy function Wfld(λ, x)
d W fld ( , x) 
W fld

d 
x
 W fld
x
d Wfld (, x)  i d  f fld dx
3-11
dx

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i
W fld

f fld  
W fld
x
x

Once we know the energy, current and more importantly force
can be calculated.
For a system with rotating mechanical terminal
x 
f fld  T fld
d Wfld (, )  i d  T fld d
T fld  
3-12
W fld ( , )


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Example 3.4:The magnetic circuit below consists of a single-coil
stator and an oval rotor. Because the air-gap is nonuniform, the
coil inductance varies with rotor angular position, measured
between the magnetic axis of the stator coil and the major axis
of the rotor, as
L( )  L0  L2 cos(2 )
where where L0=10.6 mH and L2=2.7 mH. Note the secondharmonic variation of inductance with rotor angle θ.
3-13
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3.5 DETERMINATION OF MAGNETIC FORCE AND TORQUE FROM
COENERGY
d Wfld (, x)  i d  f fld dx
Mathematically manipulated to define a new state function
known as the COENERGY, from which force can be obtained
directly as a function of current.
 (i, x)  i   W fld (, x)
W fld
 (i, x)   di  f fld dx
dW fld
Note that energy and coenergy
equal for linear systems.
3-14
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 (i, x) 
d W fld

W fld
i
di 
x

 W fld
x
dx
i
 (i, x)   di  f fld dx
dW fld

 (i, x)
W fld
i
f fld 
 (i, x)
W fld
x
i
 (i, x)    (i, x) di
W fld
0
3-15
x
i
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In field-theory terms, for soft magnetic materals
(B=0 when H=0)
 H0

     B dH  dV
W fld


V 0

For permanent magnet materials (B=0 when H=Hc)
 H0

     B dH  dV
W fld


V  Hc

3-16
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Effect of x on the energy and coenergy of a singly-excited device: (a)
change of energy with  held constant; (b) change of coenergy with i held
constant. Figure 3.11
3-17
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Example 3.5: For the relay below, find the force on the plunger as a
function of x when the coil is driven by a controller which
produces a current as a function of x of the form
x
i ( x)  I 0   A
d 
3-18
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Example 3.6: The magnetic circuit in the figure is made of high-permeability
electrical steel. The rotor is free to turn about a vertical axis. The dimensions
are shown in the figure.
a)
Derive an expression for the torque acting on the rotor in terms of the
dimensions and the magnetic field in the two air gaps. Assume the reluctance
of the steel to be negligible and neglect the effects of fringing.
b)
The maximum flux density in the overlapping portions of the air gaps is to be
limited to approximately 1.65 T to avoid excessive saturation of the steel.
Compute the maximum torque for r1=2.5 cm, h=1.8 cm, and g=3 mm.
3-19
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3.6 MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS
Many electromechanical devices have multiple electrical terminals.
3-20
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USING ENERGY FUNCTON:
d W fld (1, 2 , )  i1 d1  i2 d2  T fld d
W fld (10 , 20 , 0 ) 
20
10
0
0
 i2 (1  0, 2 ,  0 ) d2   i1 (1, 2  20 ,  0 ) d1
For magnetically linear systems
1  L11 i1  L12 i2
2  L21 i1  L22 i2
L22 1  L12 2
i1 
D
 L21 1  L11 2
i2 
D
D  L11L22  L12L21
3-21
Integration path to obtain Wfld(10, 20, 0).
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
W fld (10 , 20 , 0 ) 
20

0
W fld (10 , 20 ,0 ) 
L11( 0 ) 2
d2 
D( 0 )

L22 ( 0 ) 1  L12 ( 0 ) 20
D( 0 )
0
d1
L11(0 ) 2 L22 (0 ) 2 L12 (0 )
20 
10 
10 20
2 D(0 )
2 D(0 )
D(0 )
T fld  
3-22
10
W fld

1 , 2
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USING COENERGY FUNCTON:
 (i1, i2 , )  1 di1  2 di2  T fld d
d Wfld
i20
i10
0
0
 (i10 , i20 , 0 )   2 (i1  0, i2 ,   0 ) di2   1 (i1 , i2  i20 ,   0 ) di1
W fld
1  L11 i1  L12 i2
2  L21 i1  L22 i2
 (i10 , i20 ,0 ) 
W fld
L11(0 ) 2 L22 (0 ) 2
i10 
i20  L12 (0 ) i10 i20
2
2
T fld 
3-23

W fld

i1 ,i2
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T fld
i12 d L11( ) i22 d L22 ( )
d L12 ( )


 i1 i2
2 d
2 d
d
For a general n electrical terminal
 1   L11 L12
   L
 2    21 L22
  

  
n   Ln1 Ln 2
 L1n   i1 
 L2 n  i2 
  
 
 Lnn  in 
1 T
  I L( ) I
W fld
2
3-24
λ  L( ) I
T fld
1 T d L( )
 I
I
2
dθ
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Example 3.7: In the figure, the inductances in henrys are given as
L11=(3+cos 2θ)x10-3; L12=0.3 cos θ; L22=30+10 cos 2θ. Find and
plot the torque Tfld(θ) for current i1=0.8 A and i2=0.01 A.
3-25
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T fld  (1.64sin 2  2.4 sin  ) 103
3-26
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3.7 FORCES AND TORQUES IN SYSTEMS WITH PERMANENT MAGNETS
Special case must be taken when dealing with hard magnetic material
because magnetic flux density is zero when H=Hc not when H=0.
•Consider fictitious winding
•In normal operation, the fictitious winding carries NO current
•Current in the winding can be adjusted to zero out the field produced by
permanent magnet in order to achieve the “zero force” starting point.
3-27
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 (i f , x)   f di f  f fld dx
dW fld
 (i f  0, x) 
W fld
   dW fld

 dW fld
path1a
x
path1b
0
0
I f0
 (i f  0, x)   f fld (i f  I f 0 , x) dx    f (i f , x) di f
W fld
If0 is the current to zero-out the field.
 (i f  0, x) 
W fld
0
  f (i f , x) di f
I f0
3-28
Integration path for calculating Wfld (if = 0, x ) in the permanent
magnet system of Fig. 3.17. Figure 3.18
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Example 3.8: The magnetic circuit
is excited by a samariumcobalt permanent magnet and
includes a movable plunger.
Also shown is the fictitous
winding of Nf turns carrying a
current if which is included
here for the sake of the
analysis. The dimensions are:
Wm=2 cm, Wg=3 cm, W0=2 cm,
d=2 cm, g0=0.2 cm, and D=3
cm.
a) Find an expression for the
coenergy of the system as a
function of plunger position x,
b) Find an expression for the
force on the plunger as a
function of x,
c) Calculate the force at x=0 and
x=0.5 cm.
3-29
Figure 3.19
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A different solution for permanent magnet circuits:
e 

   R A   H c  
d 

 ( Ni )eq e 
   R A 
 
d 
 d
( Ni)eq  Hc d
3-30
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Example 3.9: Figure shows an actuator consisting of an infinitely-permeable yoke
and plunger, excited by a section of NdFeB magnet and an excitation winding
of N1=1500 turns. The dimensions are: W=4 cm, W1=4.5 cm, D=3.5 cm, d=8 mm,
and g0=1 mm.
a)
Find x-directed force on the plunger when the current in the excitation winding
is zero and x=3 mm.
b)
Calculate the current in the excitation winding required to reduce the plunger
force to zero.
3-31
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3.8 DYNAMIC EQUATIONS
We are interested in the operation of complete electromechanical system and not just of the
electromechanical energy conversion system around which it is built.
For Electrical Terminal:
di
d L( x ) d x
v0  R i  L( x)  i
dt
d x dt
For multiple-excited system, we will have similar equation for each terminal
3-32
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For Mechanical Terminal:
Spring:
f K  K ( x  x0 )
K : Spring constant
(N/m)
x
K
Damper:
dx
f D  B
dt
B
: Damping constant (N.s/m)
B
Mass:
f M  M
f fld
3-33
f fld
2
d x
d t2
M : Mass
(kg)
d 2x
dx
M 2 B
 K ( x  x0 )  f 0
dt
dt
M
f0
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Dynamic Equations (Electrical and Mechanical Equations Together):
d i(t ) d L( x) d x(t )
v0 (t )  R i(t )  L( x)
i
dt
dx
dt
d 2 x(t )
d x(t )
f 0 (t )  M

B
 K ( x(t )  x0 )  f fld (i(t ), x(t ))
2
dt
dt
f fld
i 2 d L( x)

2 dx
These equations completely specify the behavior of electromechanical device. Solution
of these equations will describe the position x and the current i at any time t in the
system.
3-34
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For Rotational Mechanical Terminal:
Torsional Spring:
TK  K (  0 )
K : Torsional Spring constant
(N.m/rad)
Friction:
d
TF   B
dt
B
: Friction constant (N.m.s/rad)
Inertia:
d 2
TJ   J 2
dt
T fld
3-35
J
: Inertia constant (kg.m2/rad)
d 2
d
 J 2 B
 K (  0 )  T0
dt
dt
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Example 3.10: Figure shows in cross
section a cylindrical solenoid magnet
in which the cylindrical plunger of
mass M moves vertically in brass
quide rings of thickness g and mean
diameter d. The permeability of brass
is µ0. The plunger is supported by a
spring with K constant. Its
unstretched length is l0. A mechanical
load force ft is applied to the plunger
from the mechanical system
connected to it. Assume that
frictional force is linearly proportional
to the velocity with coefficient B. The
coil has N turns and resistance R. Its
terminal voltage is vt and its current i.
Derive the dynamic equations of
motion of the electromechanical
system.
3-36
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EXTRA Example: A two poles VR machine is shown in figure. Stator and rotor has
infinite permeability.
a)
Find gap cross-sectional area as a function of θ.
b)
Find the inductance for the machine.
c)
Write down the dynamic equations.
d)
Solve the dynamic equations to find the position of rotor as a function of time
initially starting from θ0=25 degrees.
Numerical Values:
N=100 turns, g=0.0005 m, d=0.1
m, r=0.04 m, J=0.05, B=0.02,
θ0=30, R=0.5 ohm, E=10 Volt.
r
0
3-37
Roto
es
r Ax

Stator Axes