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ELX 311 Chapter 1
Magnetic Circuits and Materials
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1.1 INTRODUCTION TO MAGNETIC CIRCUITS
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1.1 INTRODUCTION TO MAGNETIC CIRCUITS
Primary way by which energy is converted from one form to another in Generators, Motors and TransformersFour basic principles of how magnetic fields are used
1. Current in a wire causes a magnetic field.2. Transformer action (Time changing magnetic field through
a coil induces a voltage in the coil)3. Motor action (Current carrying wire in a magnetic field has
a force induced on it)4. Generator action (Moving wire in a magnetic field has a
voltage induced in it)
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1.1 Intro to Magnetic Circuits (2)
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How a coil creates a magnetic field
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1.1 Intro to Magnetic Circuits (3)
DEFINITIONSH: Magnetic Field Intensity. A measure of the effort that a current is putting into the establishment of a magnetic field. [At/m] = Ampereturns / meter: Magnetic Permeability of material. The relative ease of establishing a magnetic field in a given material. [H/m] = Henry / meterB: Magnetic Flux Density. Measure of the number of flux lines per unit area. [Wb/m2] = Weber / square meter or [T] = Tesla
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1.1 Intro to Magnetic Circuits (4)
Amperes Law
0
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The magnetomotive force (mmf )in ampere-turns around any closed path is equal to the net current in amperes enclosed by the path.
Value of H is the component of
H tangent to ds.
Hx
Hx
Hx
1.1 Intro to Magnetic Circuits (5)
Amperes Law
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1.1 Intro to Magnetic Circuits (6)
Magnetic circuit
Valid Assumptions:All the current remains within the conductorAll Magnetic Flux () remain within the high permeability
magnetic core, i.e. negligible leakage fluxThe cross sectional area Ac is perpendicular to the flux lines
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1.1 Intro to Magnetic Circuits (7)
Magnetic circuit
because path length of any flux line is close to the mean core length lc
Since = Ni Hclc = Ni = F Or
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1.1 Intro to Magnetic Circuits (8)
Permeability and Flux
= 0r where:0: Permeability of free space ( air)
0 = 4pi 10-7 H/mr: Relative permeability of the specific material
r range from 2000 80000: Flux in Weber [Wb]
= BAA = perpendicular cross sectional area through which flux lines are cutting
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1.1 Intro to Magnetic Circuits (9)
Magnetic Circuit with air gap
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1.1 Intro to Magnetic Circuits (10)
Magnetic Circuit with air gap
RRRR: Reluctance in [At/Wb]
: Flux in [Wb]
FFFF: Magnetomotive Force (MMF) in [At]
= Ni
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1.1 Intro to Magnetic Circuits (11)
Electrical Circuit vs Magnetic Circuit
= Ni
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1.1 Intro to Magnetic Circuits (11b)
Polarity of the MMF
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1.1 Intro to Magnetic Circuits (8)
Magnetic Circuits
Magnetic Circuit Calculations are always an approximation because of:1. Leakage flux2. Assumption of a mean path length3. Nonlinearity of since = f() (See Sec. 1.3)4. Fringing (Ignored in this book)
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Example 1-1
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Example 1-1
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Do:Example 1.2 our way
Practice Problems 1.1 and 1.2
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1.2 FLUX LINKAGE, INDUCTANCE AND
ENERGY
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1.2 Flux linkage, inductance and energy (1)
Faradays Law
!" # $$& #$$&
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Problem with Faradays equation It assumes that the same amount of flux is present in each
turn of the coil.
NOT the case due to flux leakage.! $$& !" *!
+, *$$&
+, $$&*
+,
!" $$& where *
+,[Wbt]
1.2 Flux linkage, inductance and energy (2)
Flux LinkageFlux that links the coils
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1.2 Flux linkage, inductance and energy (3)
Inductance
If {(r = constant) or (RRRRgggg >> RRRRcccc )} thenRelation between and i = linear
Inductance:
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1.2 Flux linkage, inductance and energy (4)
Inductance
Inductance is measure in Henrys (H)
or weber-turns per ampere
Look at proportionality of Inductance for the case (RRRRgggg >> RRRRcccc )} :
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1.2 Flux linkage, inductance and energy (5)
Example 1.3
Given: N-turns, current = i, r =, g1 , A1, g2 , A2.Find: a) L b) B1
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1.2 Flux linkage, inductance and energy (6)
Example 1.4
Given: Based on ex. 1.1, r = 72300 2900Find: Inductance for two r values.
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1.2 Flux linkage, inductance and energy (7)
Inductance (Self and Mutual)
Resultant Core Flux:
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1.2 Flux linkage, inductance and energy (8)
Inductance (Self and Mutual)
Flux linkage in coil 1 due to i1 and i2 respectively:
Self Inductance
Mutual Inductance
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1.2 Flux linkage, inductance and energy (9)
Energy
For static magnetic circuit
Power [W= J/s]:
W [J]:
1.2 Flux linkage, inductance and energy (10)
Energy
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1.3 PROPERTIES OF MAGNETIC MATERIALS
1.3 Properties of Magnetic Materials (1)
Importance of Magnetic Materials
Magnetic Material are important in the context of
electromechanical energy conversion devices:
1. Large B with low F - high B means high energy density. Like choosing between different types of conductors,
certain conductors conduct better than other.
2. MM constrain and direct magnetic fields in well-defined
paths
Transformers: maximise coupling and lower excitation
current.
Rotating Machines: shape the fields to obtain desired
torque-production and electrical terminal
characteristics.
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1.3 Properties of Magnetic Materials (2)
Magnetic Domains
Iron (Fe = Ferrous) and alloys of iron with other metals
(cobalt, tungsten, nickel, aluminium)
http://mujiholic-technoholic.blogspot.com/2008/01/do-
you-know-magnet-works.html
Magnetic Domains:
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/ferro.html
1.3 Properties of Magnetic Materials (3)
Hysteresis and the B-H Curve
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/hyst.html
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35B-H loops for M-5 grain-oriented electrical steel 0.012 in thick. Only the top halves of the loops are shown here. (Armco Inc.) Figure 1.9
1.3 Properties of Magnetic Materials (4)
Hysteresis and the B-H Curve
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1.3 Properties of Magnetic Materials (5)
DC or Normal Magnetization Curve
http://info.ee.surrey.ac.uk/Workshop/advice/coils/BHCkt/index.html
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37B-H loops for M-5 grain-oriented electrical steel 0.012 in thick. Only the top halves of the loops are shown here. (Armco Inc.) Figure 1.9
1.3 Properties of Magnetic Materials (6)
DC or Normal Magnetization Curve
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1.3 Properties of Magnetic Materials (7)
DC or Normal Magnetization Curve
Dc magnetization curve for M-5 grain-oriented electrical steel 0.012 in thick. (Armco Inc.)Figure 1.10
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1.3 Properties of Magnetic Materials (8)
Practice Problem 1.6
1.6 T
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1.4
AC EXCITATION
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1.4 AC Excitation (1)
Induced Voltage
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1.4 AC Excitation (2)
Induced Flux and Voltage How??
http://skelectricals1997.hpage.co.in/ac_alternator_-_basic_1802439.html
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1.4 AC Excitation (3)
Excitation Current i
B = /A and H = Ni/l, Hence B-H curve can be replace by a -I curve
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1.4 AC Excitation (4)
Magnetisation Current
From: Electric Machinery Fundamentals, SJ Chapman, 4ed
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1.4 AC Excitation (5)
Excitation rms VAsAC excitation characteristic of core materials
VAs required to excite a core:
Core
VolumeExcitation rms Vas pu mass:
1.4 AC Excitation (6)
Excitation rms VAsAC excitation characteristic of core materials
Normalised rms excitation VAs a material property, ie
independent of shape and size
Depends only on Bmax because Hrms = function of Bmaxbased on BH curve at a specific frequency.
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1.4 AC Excitation (7)
Excitation rms required for a magnetic material per unit weight
Exciting rms voltamperes per kilogram at 60 Hz for M-5 grain-oriented electrical steel 0.012 in thick. (Armco Inc.)Figure 1.12
1.4 AC Excitation (8)
Excitation Current ComponentsThe excitation current supplies:
a. the mmf required to produce the core flux
(magnetization current) and
b. the power input associated with the energy in
the magnetic field in the core.
POWER
Active Reactive
Core (Heat)
Losses
Eddy
CurrentsHysteresis
Magne-
tization
iex = icore_loss + imagnetization
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1.4 AC Excitation (9)
Eddy Current Losses
Peddy = k B2max f
2
1.4 AC Excitation (10)
Hysteresis Losses
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1.4 AC Excitation (11)
Pcore = Peddy + Physteresis
Core loss at 60 Hz in watts per kilogram for M-5 grain-oriented electrical steel 0.012 in thick. (Armco Inc.) Figure 1.14
Core loss depend on:
Metallurgy of the material, ie type of material
Flux density
Frequency
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1.4 AC Excitation (12)
Grain-Oriented (Silicon) Steel
Nonoriented Steel Oriented Steel
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1.4 AC Excitation (13)
Grain-Oriented Silicon Steel
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1.4 AC Excitation (14)
Grain-Oriented Silicon Steel
www.learnabout-electronics.org
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1.4 AC Excitation (15)
Example 1.8
1 in = 2.54 cm
Mean flux
path length lc
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1.5 PERMANENT MAGNETS
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1.5 Permanent Magnets (1)
Residual flux density Br and coercivity Hc
1.5 Permanent Magnets (2)
Residual flux density Br and coercivity Hc
(a) Second quadrant of hysteresis loop for Alnico 5;(b) second quadrant of hysteresis loop for M-5 electrical steel; (c) hysteresis loop for M-5 electrical steel expanded for small B. (Armco Inc.)Figure 1.16
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1.5 Permanent Magnets (3)
Example 1.9
0.25
0.3
1.5 Permanent Magnets (4)
Coercivity Hc
Coercivity Hc is a measure of: the magnitude of the mmf required to demagnetize the
material
The capability of the material to produce flus in a
magnetic circuit which includes an air gap (ex 1.9)
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1.5 Permanent Magnets (4)
Maximum Energy Product
Useful measure of the
capability of a PM
material.
Corresponds to the
largest B-H product,
which in turn corres-
ponds to a point on the
second quadrant of the
hysteresis loop.
This point results in the
smallest volume of that
material required to
produce a given flux
density in an air gap.
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1.5 Permanent Magnets (5)
Maximum Energy Product
[Joule]
[m3] [J/m3]
Choosing a material with the largest available
maximum energy product can result in the smallest
required magnet volume.
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1.5 Permanent Magnets (6)
Maximum Energy ProductObjective: Find an equation that will give me the volume of
the magnetic material required to produce a specific
magnetic flux density inside a given sized airgap.
From Example 1.9:
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1.5 Permanent Magnets (7)
Maximum Energy ProductObjective: Find an equation that will give me the volume of
the magnetic material required to produce a specific
magnetic flux density inside a given sized airgap.
1. Minimize Vm by operating
the magnet at point of
maximum energy product, ie
the load line must go
through (HmBm)max
2. The larger (HmBm), the
smaller the required magnet
to produce the required Bg.
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1.5 Permanent Magnets (8)
Example 1.10
Find: The Vg,min for Bg = 0.8 T
Step 1: Graphically estimate the point of maximum energy product on
the BH curve.
Pt1: H = -30, B 1.1 HB = -33 kJ/m3
Pt2: H -40, B 1.0 HB = -40 kJ/m3
Pt3: H -45, B 0.8 HB = -36 kJ/m3
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1.5 Permanent Magnets (9)
Example 1.10
Find: The Vg,min for Bg = 0.8 T
Step 2: From Eq 1.57 calculate Am:
= AmBm = AgBg so
Am = AgBg/ Bm= 2cm20.8/1
= 1.6cm2Step 3: From Eq 1.58 calculate lm:
F = 0 = Hmlm + Hglg so
lm = -Hglg/ Hm = -Bglg/0Hm= -0.80.2cm/(4pi10-7 -40 103)= 3.18 cm
Step 4: Hence the minimum magnet volume required is:
Vm = lm Am= 3.18 x 1.6
= 5.09 cm3