ECE 325 – Electric Energy System Components
3- Fundamentals of Electromagnetism and
Rotational Motion
Instructor:
Kai Sun
Fall 2015
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Content
• Fundamentals of electromagnetism (Ch. 2.16-2.31)
• Fundamentals of rotational motion (Ch. 3.0-3.14)
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Magnetic Flux Density
• Defined as the amount of magnetic flux in an
area taken perpendicular to the magnetic
flux’s direction.
• It is a vector 𝐵, having its magnitude and
direction at each point in space.
• This course only concerns its magnitude B.
• Average B=  /A (unit: T=Wb/m2)
where  - magnetic flux
A - cross section area
(perpendicular to the flux)
B
A
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Magnetic Field Strength (Intensity)
• Denoted by H (A/m), which is
proportional to B
• B-H curve
– Vacuum and nonmagnetic materials
B=0H
0 - magnetic constant or permeability
of vacuum
– Magnetic material
B= r0H (with saturation)
Slope=r0
r - relative permeability
(not constant)
4
Ampere’s Circuital Law
• The law states the relationship between the integrated
magnetic field around a closed loop and the electric
current passing through the loop.
Fm  N  I 
 H  dl  H  l
I
l
Fm: magnetomotive force (mmf) in A or Aturns
Hl : magnetic potential, analogous to the electric potential.

N
I
N
H
l0
l
l
I
H
Analogous to KVL for an electrical circuit.
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Magnetic Reluctance (Resistance)
• Hopkinson’s law (a counterpart to Ohm’s law):
Fm=Rm
Rm- Magnetic reluctance (unit: Aturns/Wb), i.e. the ratio of the
mmf in a magnetic circuit and the magnetic flux in this circuit.
(Note: it is not a constant.)
Fm=NI=Hl=Rm=BARm=r0HARm
Rm 
l
(=r0: permeability)
Analogous to the resistance in an electrical circuit
 r 0 A
l
R
A
(: resistivity)
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Example 1
I
The closed core is made of even magnetic material, and
has flux leakage. Its average magnetic filed line has
length l=0.45 m, and exciting winding has 300 turns.
Calculate the current I to have magnetic flux density
B=1.4 T respectively for two magnetic materials:
• silicon iron (1%)
• cast steal.
l
1 =0.0014 2 =0.0007
Solution
From Figure 2.26, H1=1000 A/m and H2=2000 A/m
H1l 1000  0.45

 1.5A
N
300
H l 2000  0.45
I2  2 
 3.0A
N
300
I1 
Observation: to have the same B, a larger current is needed for a
material with lower permeability.
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Example 2
The toroid core made of magnetic material has an inner diameter of 10 cm and an outer
diameter of 15 cm. There is an air gap of =0.2 cm wide. The current I in the conductor
is 1 A. To have magnetic flux density B=0.9 T, how many turns do we need for the
exciting winding? Assume that the magnetic material has H=500 A/m at B=0.9T.
Solution
The magnetic field strength of the air gap:
The total average length of magnetic field lines:
l    (0.1  0.15) / 2  0.393m
The average length of magnetic filed lines in the core:
l1=l -  =0.393-0.002=0.391m
The total mmf
Fm=H0 + H1l1 = 7.2105 0.002 + 500 0.391
=1440 + 195.5=1635.5 (Aturns)
The turns of the current: The air gap has much smaller permeability, and hence much
N=Fm/I=1636 (turns) larger magnetic resistance, so the magnetic potential drop across
the air gap is much larger than that in the magnetic material.8
Faraday’s Law of Electromagnetic Induction
1. If the flux linking a loop (or turn) varies as a function of time, a
voltage is induced between its terminals.
2. The value of the induced voltage is proportional to the rate of
change of flux.
E  2000 
0.002  0.005
0.1
 60V
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Voltage Induced in a Conductor
• Whenever a conductor (part of a
moving coil) cuts a magnetic field B,
a voltage E is induced across its
terminals, which is proportional to its
active length l and relative speed v.
E=Blv
E
l
d d ( BA) BdA Bldx



 Blv
dt
dt
dt
dt
• The direction of the current in the
conductor follows the right-hand rule.
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Example
• B=0.6T, l=2m, v=100m/s.
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Lorentz Force on a Conductor
• Also called electromagnetic force, it is the force a current-carrying
conductor is subjected to when it is placed in a magnetic field.
F=BlI (unit: N)
• The direction of the Lorentz force follows the left-hand rule
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Hysteresis Loop and Hysteresis Losses
Hysteresis loop
Magnetic domains oriented
randomly
Magnetic domains lined up in
the presence of an external
magnetic field.
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Eddy Currents and Eddy losses
• The cause of eddy currents by an AC flux 
• How to reduce eddy currents
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Comparison of Electrical and Magnetic Circuits

Magnetic
circuit
mmf
I
Hl
N
Fm=NI
Magnetic
flux

I
Electrical
circuit
emf
+
E
_
U
Magnetic
potential
Hl
Current
Electric
potential
I
U
R
E
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Comparison of Electrical and Magnetic Circuits
Magnetic
circuit
Hopkinson’s
Law

Magnetic Magnetic
resistance
flux
density
Ampere’s
Circuital
Law
Gauss’s
Law
KVL
KCL
I
N
Nl
Electrical
circuit
Ohm’s Law
I
+
_
E U
R
E
I
R
Electrical
resistance
Current
density
l
R
A
I
J
A
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Rotational Motion of Electric Machines
• An electric machine rotates about a
fixed axis, called the shaft, so its
rotation is restricted to one angular
dimension.
• Relative to a given end of the
machine’s shaft, the direction of
counterclockwise (CCW) rotation
is often assumed to be positive.
• Therefore, for rotation about a
fixed shaft, all the concepts are
scalars.
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Angular Position, Velocity and
Acceleration
• Angular position 
– The angle at which an object is
oriented, measured from some
arbitrary reference point
– Unit: rad or deg
– Analogy of the linear concept • Angular acceleration  =d/dt
of distance along a line.
– The rate of change in angular
velocity with respect to time
• Angular velocity =d/dt
– Unit: rad/s2
– The rate of change in angular
position with respect to time
•  and  >0 if the rotation is CCW
– Unit: rad/s or r/min (revolutions •  >0 if the absolute angular
per minute or rpm for short)
velocity is increasing in the CCW
direction or decreasing in the CW
– Analogy of the concept of
direction
velocity on a straight line.
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Moment of Inertia (or Inertia)
•
•
Inertia depends on the mass and shape of the object (unit: kgm2)
A complex shape can be broken up into 2 or more of simple shapes
m
J  ( R12  R22  R1 R2 )
3
J
mL2
J
12
m 2
( R1  R22 )
2
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Torque and Change in Speed
• Torque is equal to the product of the force times the perpendicular distance
between the axis of rotation and the point of application of the force.
T=Fr (Nm)
T=0
T
T=Fr
• Newton’s Law of Rotation: Describes the relationship between the total
torque applied to an object and its resulting angular acceleration.

J n
T  J  J


t 9.55 t
60[min] 30
n[r/min]  [rad/s] 
   9.55
2

n: change in speed (in r/min or rpm)
t: interval of time during which the torque is applied (in second)
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Mechanical Work, Power and Kinetic Energy of
Rotational Motion
• Applying a constant torque T (Nm) to a motor
– Work:
W=T (J)
: angular distance (rad)
– Power:
Tn
P  T 
9.55
(W)
: angular velocity or speed of rotation (rad/s)
n: speed of rotation (r/min)
2
1

2
Jn 2  5.48 103 Jn 2 (J)
– Kinetic Energy: Ek  J  
2
1800
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Example
A solid 1400 kg steel flywheel has diameter of 1 m
and a thickness of 225 mm. Calculate:
• Its moment of inertia
• To increase its speed from 60 r/min to 600 r/min
by applying a torque of 20 Nm. For how long must the torque be applied?
• The kinetic energy when the flywheel revolves at 1800r/min
Solution:
Ignoring the thickness of the flywheel, the moment of inertia:
J n 175 600  60
t 


 494.7 s
9.55 T 9.55
20
J n
T

9.55 t
Ek  5.48 103 Jn 2 =5.48 103 175 18002 =3.1MJ
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Mechanically Coupled System
• Assume that under normal operating conditions, TM (or n) is cw and TL is ccw.
• When TM=TL and have opposite directions, speed n (or ) of the shaft is constant
(no matter it is cw, ccw or even 0)
• In order to increase n to n1 (cw) • In order to decrease & reverse n to n2 (ccw)
– current I   TM<TL n 0 ccw
– current I   TM>TL n
– once n=n2, I   TM=TL
– once n=n , I   T =T
1
M
L
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Conclusions on a Mechanically Coupled System
• Speed:
– When TM=TL and have opposite
directions, the actual steady-state speed
being cw (or ccw) depends on whether
TM was greater (or less) than TL for a
certain period of time before the actual
steady-state condition was reached.
• Power flow:
– When TM and n have the same direction
(i.e. P>0), the motor delivers power to
the load (in the motor mode); otherwise,
the motor receives power from the load
(in the generator mode)
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Question 3-11
A motor develops a cw torque of 60Nm, and the load develops a
ccw torque of 50Nm.
• If this situation persists for some time, will the direction of
rotation eventually be cw or ccw?
cw since 60Nm (cw) > 50Nm (ccw)
• What value of motor torque is needed to keep the speed
constant?
50Nm
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Electric Motors Driving Linear Motion Loads
• Power output
Po=Fv
F in N and v in m/s
• Power input
Pi=T n/9.55
• Assuming no losses
Pi=Po
T n=9.55F v
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Homework #2
• Read Ch. 2.16-2.31, 3.0-3.14
• Questions 2-5, 2-8, 2-9, 3-12, 3-17, 3-18, 3-19
• Due date: 9/23 (Wed)
– hand in your solution in the class or
– to Denis at MK 205 or by email (dosipov@vols.utk.edu)
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