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EEE336 Electromechanical
Energy Conversion
Dr. Bülent Dağ
1
Reference Books
Textbook:
S.J. Chapman, Electric Machinery Fundamentals,
McGraw Hill
Supplementary book:
A.E. Fitzgerald, et.al., Electric Machinery, McGraw Hill,
6th ed.
2
Office Hours
Tuesday: 10:30-12:30, Friday: 10:3012:30
Office #: 323
E-mail: bulentdag@gazi.edu.tr
3
Assesment Criteria
Exam
Number
Weight
Midterm
1
% 40
Final
1
% 60
4
Content of the Lecture
• Electromagnetic circuits and energy conversion
principles
• Transformers
• DC Machines
– DC machine fundamentals
– DC motors
– DC generators
• AC Machines
– AC machine fundamentals
– Three-phase Induction motors
– Synchronous generators
5
Lecture 1
Introduction To Machinery Principles and
Magnetic Circuits
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Energy Conversion with Electrical
Machines
Motors
Mechanical Energy
Electrical Energy
Generators
Types of electrical machines:
• DC Machines
• AC Machines
 Induction machines
 Synchronous machines
• Special machines (step motor, reluctance machine, BDCM)
7
Electrical Machines
Definitions:
• Shaft: The axis of rotation in the electric machines is called shaft.
Direction of rotation can be either clock wise (CW) or counter
clock wise (CCW). CCW will be assumed positive in this course.
• Angular position (θ): The angle of the rotating object with respect to an
arbitrary reference point. The unit is radians or degrees. Analogues
to distance (x) in linear motion.
• Angular velocity (ω): Angular velocity (speed) is the time rate at which the
angular position (θ) changes. It is analogous to linear velocity (v).
• Angular acceleration (α): Rate of change of angular velocity with respect
to time.
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Rotational Motion
• m , Angular Velocity, radians/second (rad/sec)
• fm, Angular Velocity, revolution/second (rev/sec)
• nm, Angular Velocity, revolution/minute (rpm)
•  , Angular Acceleration, radians/second2
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
9
• Torque (T): Analogous to force (F) in linear motion
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Newton’s Law of Motion
Linear motion
Rotational motion
J: Moment of inertia of the rotating object, kg-m2
Work:
Linear motion
Rotational motion
If F is constant, W = Fx, (Joules)
Power:
Linear motion
If T is constant, W = Tθ, (Joules)
(Joules/second = Watts)
Rotational motion
11
Maxwell’s Equations
Gauss’s Law
Faraday’s Law
Ampere’s Law
12
Electromagnetic Systems
λ
Faraday’s law
Ampere’s law
Lorentz’s laws
13
The Magnetic Field
•
Magnetic field intensity H (A-turns/m) is governed by Gauss
and Ampere’s Law
(Gauss Law),
•
(Ampere Law)
Magnetic Flux density, B (Tesla or T) depends on the medium
property and related to H as follows,
Where,
µ is magnetic permeability of medium
µo is permeability of the free space =
µr is the relative permeability of the material =
14
•
Magnetic Flux density B (Tesla or T) accounts for the magnetic
prosperities of the medium.
• B vs H relationship is frequently expressed by a non-linear curve called
B-H curve, as shown in figures below.
• In most ferromagnetic materials, the curve starts at a very high slope
which tends to be constant. This is the linear portion of the B-H curve.
• At higher values of H, the flux density levels off and the material is said
to be in saturation region.
15
A simple magnetic core
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Magnetic Circuit
(Weber, Wb)
17
Magnetic Circuits
Flux equation can be rewritten as follows,
Define magnetomotive force, F and reluctance, R
(A-turns, A-t),
(A-t/Wb),
Then basic magnetic circuit equation in correlation with electric circuits,
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Magnetic Circuits
The reciprocal of the reluctance is called permeance, Λ
For series connected parts;
Polarity of mmf, F
For parallel connected parts;
Assumptions made in magnetic circuit
analysis;
1. Mean core length and uniform flux
density
2. No leakage flux
3. Constant permeability, µ
4. No flux fringing
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Fringing and leakage flux
leakage
fringing
Fringing: When there exists an air gap, flux may
not follow the shortest path while jumping from
one part to the other part, especially at the edges.
As a result, effective cross sectional area of the
air gap increases.
Leakage: Some of the flux produced complete its
turn around the winding through the air.
20
Example (1.1-Chapman)
A ferromagnetic core is shown in
the Figure. Three sides of this
core are of uniform width while
the fourth side is somewhat
thinner. The depth of the core
(into the page) is 10 cm. and the
other dimensions are shown in
the figure. There is a 200 turn
coil wrapped around the left side
of the core. Assumming relative
permeability µr of 2500, how
much flux will be produced by a
1 A input current?
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Solution (1.1-Chapman)
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Example (1.2-Chapman)
A magnetic core with a relative
permeability of 4000 is shown in the
Figure. Assuming a fringing coefficient of
1.05 (or %5 increase in airgap cros
sectional area) for the air gap, find
a) total (equivalent) reluctance of the
system
b) the flux density in the air gap if i = 0.6 A.
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The magnetic circuit is shown in figure (b).
0.0012
0.0012
0.0012
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Example (1.3-Chapman)
Figure (a) shows a simplified rotor and
stator for a dc motor. The mean path
length of the stator is 50 cm, and its crosssectional area is 12 cm2. The mean path
length of the rotor is 5 cm, and its crosssectional area also may be assumed to be
12 cm2. Each air gap between the rotor
and the stator is 0.05 cm wide, and the
cross-sectional area of each air gap
(including fringing) is 14 cm2. The iron of
the core has a relative permeability of
2000, and there are 200 turns of wire on
the core. If the current in the wire is
adjusted to be 1 A, what will the resulting
flux density in the air gaps be?
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Solution (1.3-Chapman)
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Magnetic Behavior of Ferromagnetic Materials
•
•
•
Constant permeability assumption is not always valid
Ferromagnetic cores have non-linear B-H characteristics
To obtain actual B-H plot, apply Vac at rated value and plot integral of Vac vs.
i on an oscilloscope in x-y mode. This plot is called magnetization curve.
Vac
Remember that,
and
where,
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Magnetic Behavior of Ferromagnetic Materials
• Therefore, reluctance of the core is not constant, actually.
• To find reluctance, we need permeability,

• To find  , first we need the operating point on B-H curve.
• If permeability is not constant then there are two solution
approaches according to given parameters;
1. If B is given and current is asked: From B-H curve determine
operating H for the core. Either by using directly Ampere’s law or by
determining permeability, µ at the operating point and using
magnetic circuits, calculate the current.
2. If current is given and B is asked: The only solution way is to obtain
a second relation between B and H of the core using Ampere’s law
for the given magnetic system. Then cross-section of B-H curve of
the core and B-H relation from the system is the operating point.
29
Example (1.4-Chapman)
Find the relative permeability of the typical ferromagnetic material whose
magnetization curve is shown in the figure below for,
(a) H = 50. (b) H = 100. (c) H = 500. and (d) H = 1000 A-turns/m
Solution:
a) H = 50 A-t/m
d) H = 1000 A-t/m
B = 0.25 T
B = 1.51 T
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Example (1.5-Chapman)
A square magnetic core has a mean path length of 55 cm and a cross-sectional
area of 150 cm2. A 200 turn coil of wire is wrapped around one leg of the core. The
core is made of a material having the same magnetization curve in previous
example.
(a) How much current is required to produce 0.012 Wb of flux in the core?
(b) What is the core's relative permeability at that current level?
(c) What is its reluctance?
Solution: The required magnetic field density, B in the core is,
, from magnetizing curve the required intensity,
A-t/m
a)
b)
c)
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Example: T magnetic core has a B-H curve as shown
below. What is the air gap flux density if the applied current
is 0.86 A? Neglect fringing and leakage.
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Solution:
From Ampere’s Law,
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Faraday’s Law: Induced voltage from a time varying
magnetic field
Faraday’s law is the basis of transformer operation. It states that if a flux passes through a turn
of a coil of wire, a voltage will be induced in the turn of wire that is directly proportional to the
rate of change in the flux with respect to time. In equation form,
If a coil has N turns and if the same flux passes through all of them, then the voltage induced
across the whole coil is given by,
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Definition: Flux linkage, λ
Definition: Inductance, L
(Henry (H) = Wb-t/A)
Example (Fitzgerald-1.3)
The magnetic circuit of Fig (a)
consists of an N-turn winding on
a magnetic core of infinite
permeability with two parallel air
gaps of lengths g1 and g1 and
areas A1 and A2, respectively.
Find, (a) the inductance of the
winding and (b) the flux density
B1 in Gap 1 when the winding is
carrying a current i. Neglect
fringing effects at the air gap.
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Solution (Fitzgerald-1.3)
a)
b)
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Two winding construction
Note that i1 and i2 are to produce flux
in the same direction in this example.
Flux linkage of coil 1
Flux linkage of coil 2
Self inductance
Mutual inductance
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Note that the resolution of the resultant flux linkages into the components produced by i1 and i2
is based on superposition of the individual effects and therefore implies a linear flux-mmf
relationship (characteristic of materials of constant permeability).
For a magnetic circuit with a single winding;
a) Then, for static systems with constant air gap and permeability (L is constant);
b) For electromechanical systems where exist moving parts (L is not constant: air gap
changes. This, in turn changes reluctance);
For systems with multiple windings; total flux linkage of each coil must be included. Then, for a
system with two coils, for example;
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Power at the terminals of a winding
[W, J/s]
Thus, the change in magnetic stored energy ΔW in the magnetic circuit in the time interval t1 to
t1 is,
[Joules]
For a single-winding system of constant inductance, the change in magnetic stored energy as
the flux level is changed from λ1 to λ1 can be written as;
The total magnetic stored energy at any given value of λ can be found from setting λ1 equal to
zero;
39
Example (Fitzgerald):
The magnetic circuit shown in the figure
has dimensions Ac = Ag = 9 cm2, g = 0.05
cm, lc = 30 cm, and N = 500 turns. Assume
µr = 70000 for core material. Find (a) the
inductance L, (b) the magnetic stored
energy W for Be = 1 T, and (c) the induced
voltage e for a 60-Hz time-varying core flux
of the form Be = 1 sin ωt T where ω = 2π60
= 377.
Solution;
a)
c)
b)
40
Properties of magnetic materials
randomly oriented dipoles
•
•
•
•
•
external field
All materials are composed of magnetic dipoles. The flexibility of magnetic dipole to rotate
around is a measure of permeability, µ
In case of an AC excitation, the dipole orientations does not follow the same path. This is
called hysteresis.
When an external field is applied to the randomly oriented dipoles and removed, some flux
remains in the material in the direction of applied field. This is called residual flux.
The required opposite mmf to make the net flux zero again, is called magnetic coercivity.
Permanent magnets (hard magnetic materials) have very high residual flux and coercivity
compared to ferromagnetic materials (soft magnetic materials).
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AC excitation
The flux is created by a current, iφ, called magnetizing current. Since,
ϕ – i curve can be obtained from B-H curve and iφ, can be obtained from ϕ – i curve as in Fig.
(a) and (b) above.
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Magnetic losses
As can be seen from Fig. a, phase of the fundamental component of the magnetizing current
is less than 90o . This means that the energy input associated with magnetizing current has
reactive and real parts.
 Real part of this energy is dissipated as losses and results in heating of the core. Two loss
mechanisms are associated with time-varying fluxes in magnetic materials; Eddy current
and Hysteresis loss.
 The reactive power is associated with energy storage in the magnetic field. This reactive
power is not dissipated in the core; it is cyclically supplied and absorbed by the excitation
source.
Eddy current losses: Time varying magnetic fields generate electrical fields and
corresponding circulating currents inside the material, as well. These currents cause ohmic I2R
losses. In order to reduce eddy current losses, thin sheets are laminated and put together to
cut eddy current paths.
Hysteresis losses: Coercive mmf causes hysteresis losses, given by,
For one cycle of variation the energy input,
Hysteresis power loss,
W/m3 , where 1.5 < n < 2.5
43
Permanent magnets
Permanent magnets are characterized by high residual flux values and high coercive force
values. They are operated in 2nd quadrant of their hysteresis loop.
2nd quadrant of Alnico-5
•
•
•
•
2nd quadrant of M-5 steel
Remanent magnetization, Br corresponds to flux density which would remain in a closed
magnetic structure without any applied mmf.
The coercivity Hc is the value of magnetif field intensity (proportional to mmf) required to
reduce magnetic flux density to zero.
Due to high remanent magnetization, in permanent magnets there is no need for current to
produce flux.
Permanent magnets are widely used in applications such as refrigerator door holders,
loudspeakers and DC motors.
44
Example (Fitzgerald):
As shown in the figure, a magnetic circuit consists of
a core of high permeability (µ→∞), an air gap of
length g = 0.2 cm, and a section of magnetic material
of length lm = 1.0 cm. The cross-sectional area of the
core and gap is equal to Am = Ag = 4 cm2. Calculate
the flux density Bg in the air gap if the magnetic
material is (a) Alnico 5 and (b) M-5 electrical steel.
Solution:
Since the core permeability is assumed infinite, H in the core is negligible. Then,
Load line equation
a) Intersection point of the load line and B-H
curve of Alnico-5 is shown in Fig. a,
b) Intersection point of the load line and B-H
curve of M-5 steel is shown in Fig. c,
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Equivalent model of permanent magnets
If B-H curve of a magnet in 2nd quadrant
has wide linear characteristics as in the
figure, then the magnet can be equivalently
represented by an mmf and a series
reluctance as in the equations below,
where, lm is mean length, and Am is effective
crossectional area of the magnet. µp is the
slope of the B-H curve as shown in the figure.
Example: Repeat the two previous example for a magnet with B-H characteristics as in the
figure above with Hc = 250 kA/m and Br = 1 T.
Solution: The equivalent magnetic circuit of the structure is shown below,
Rm
Fm
Rg
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Magnetization curves for common permanent-magnet
materials
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