Topic 1 : Magnetic Concept and Transformer
Dr. Mohd Hafiz Habibuddin(adapted from the slides of Dr. Junaidi and Dr. Norasiah)
MAGNETISM
Magnetism is a force field that acts on some materials but not on other
materials
Magnetic field
Shape of steel fillings on a sheet over a magnet are shown below.
Magnetic field
The shape and density of the chains formed enable us to form a mental picture of a magnetic field and lead to the idea of magnetic flux lines.The density of the lines depends on the strength of the magnet. They become thinner further away from the magnet.Magnetic flux are purely imaginary lines used to visualize the distribution and density of a magnetic field.
Magnetic field
Magnetic field around a bar magnetField forms closed “flux lines” around the magnetUse compass to map out magnetic fieldTwo “poles” dictated by direction of the field – N and SMagnetic flux,Φ is measured in Webers (Wb).
Characteristic of magnetic flux
The direction of flux is from North to South.Each line of magnetic flux forms a closed loop.Lines of electric flux never intersect.Lines of magnetic flux are like stretched elastic cords, always trying to shorten themselves.Lines of magnetic flux which are parallell and in the same direction repel one another.
Magnetic field of a current‐carrying conductor
Straight conductor.The field can be described by:
right handscrew rule.
Magnetic field of a current carrying conductor
Straight conductor
Magnetic field of a current carrying conductor
Coil
Magnetic field of a current carrying conductor
CoilPlacing a ferrous material inside the coil increases the magnetic field.Acts to concentrate the field.Also notice field lines are parallel inside ferrous element.‘flux density’ has increased
Force on current carrying conductor
Flux of magnet and conductor
Interaction of flux produces force
S
N
Force
S
N
Force on current carrying conductor
Left‐hand rule
Index fingerFLUX
ThumbFORCE
Middle fingerCURRENT
Electromagnetic induction
Direction of induced emf.Fleming’s right‐hand rule.
Lenz’s law The direction of an induced emf is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that emf.
ThumbFORCE
Middle fingerCURRENT
Index fingerFLUKS
MAGNETIC CIRCUIT
Basic Definitions
Basic Definitions
Example 1
Circumference = 20cmCoil has 25 turnsCurrent required to produce magnetic field strength of 800 A‐L/m?
Solution
I = Hl / N=800 × 20 × 10‐2 / 25 = 6.4 A
Basic Definitions
Basic Definitions
Basic Definitions
For a fixed magnetic field strength, a higher permeability will gives higher flux density‐ ie. a small current can produce a large flux density
Low μ High μ
Basic Definitions
Basic Definitions
Permeability of different materialsFree space, electrical conductors (aluminium or copper), insulators: μr = unity.
Ferromagnetic materials (iron, cobalt and nickel):μr = several hundred ‐ several thousand
Notice the nonlinearity and saturation
Example 2
A magnetic material has a magnetic field strength of 3000 A‐T/m and relative permeability of 160. Determine the total flux on the material if its cross‐sectional is 5 cm2.
B = μr μo H= 160 × 4π × 10‐7, × 3000= 0.6 T
Total flux , Φ = A× B= 5 × 10‐2= 0.03 Wb
Basic Definitions
Comparison of electric and magnetic circuit
Electric circuit Magnetic circuitEMF, v MMF, NIElectric field strength, v/m
Magnetic field strength, H
Current, I Flux, ΦCurrent density, I/A Flux density, B
Comparison of electric and magnetic circuit
Since the magnetic circuit behaves in similar ways as an electric circuit, the laws of electric circuit are applicable to the magnetic circuit.Ohm’s law equivalent :
F = ΦS.
Kirchoff law equivalent:Φin = ΦoutTotal NI = Total Hl
A magnetic circuit can be redrawn as electric circuit equivalent to facilitate the analysis.
Comparison of electric and magnetic circuit
Equivalent circuit – analogy between magnetic circuit and electric circuit
Example 3
A ferromagnetic core is shown in figure below . The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 1000.
Example 3
There are three regions in this core. The top and bottom form one region, the left side forms a second region, and the right side forms a third region. If we assume mean path length of the flux is in the center of each leg of the core, and if we ignore spreading at the corners of the core, then the path length of each regions are:
( ) cml and,cml,cmcm.l 3 3030555272 21 ====
Example 3
The reluctance of each regions will be
The total reluctance is thus
( )( )( )( ) WbAt k.m.m.m/H
m.A
lAlS
ro
36581500501041000
5507
1
1
1
11 =
×=== −πμμμ
( )( )( )( ) WbAt k.m.m.m/H
m.A
lAlS
ro
7547100501041000
307
2
2
2
22 =
×=== −πμμμ
( )( )( )( ) WbAt k.m.m.m/H
m.A
lAlS
ro
49950500501041000
307
3
3
3
33 =
×=== −πμμμ
WbAtk....SSSSTotal 6201499575473658321 =++=++=
Example 3
Magnetomotive force required to produce a flux of 0.005 Wb is
The required current is
( )( ) t.AWbAtk.Wb.SF Total 100862010050 ==Φ=
A.tAt
NFI 0162
5001008
===
Example 3
The flux density on the top of the core is
The flux density on the right side of the core is
The flux density on the right side of the core isDIY
( )( ) T.m.m.
Wb.A
B 670050150
0050==
Φ=
( )( ) T.m.m.
Wb.A
B 02050050
0050==
Φ=
Magnetic Leakage
Flux not passing through in the magnetic material or in air gapOccurs at the magnetic source
Flux produced by the coil :total flux
Flux passing through magnetic material : useful fluxLeakage flux :total flux – useful flux
.leakage _ factor,α =totalflux
usefulflux
Fringing Effect
Occurs at the air gapFlux tends to bulge outwards
Increase the effective areaFlux remains the same (still a useful flux)
Reduce the flux density
Series Magnetic Circuit(with air gap)
Magnetic leakage and fringing effect are ignored in this course.
g
gg
c
cc
ggccgC
g0
gg
cc
cc
AB;
AB
densityFlux
lHlHNiSS
Ni
Al
S;A
lS
Φ=
Φ=
+=→+
=Φ
μ=
μ=
Series Magnetic Circuit(composite materials)
bS
aS
cS
ccbbaacba
cc
cc
bb
bb
aa
aa
lHlHlHNiSSS
NiA
lSA
lSA
lS
++=→++
=Φ
===μμμ
;;
Example 4
A coil of 200 turns is wound uniformly over a wooden ring having a mean circumference of 600mm and a uniform cross‐sectional area of 500mm2. if the current through the coil is 4A, calculate
the magnetic field strength(1330A/m)
the flux density(1680µT,)
the total flux(0.838µWb)
Example 5
Calculate the magnetomotive force required to produce a flux of 0.015Wb across an air gap 2.5mm long, having effective area of 200cm2
(1492At)
Example 6
A mild‐steel ring having a cross‐ sectional area of 500 mm2 and a mean circumference 0f 400mm has a coil 0f 200 turns wound uniformly around it. The relative permeability of the mild steel for the respective flux density is about 380. Calculate
the reluctance of the ringthe current required to produce a flux of 800µWb in the ring
(1.68 x 106 At/Wb, 6.7A)
Example 7
The Figure represents the magnetic circuit of a relay. The coil has 500 turns and the mean core path is lc = 360 mm. When the air‐gap lengths are 1.5 mm each, a flux density of 0.8 Tesla is required to actuate the relay. The core is cast steel with the field intensity 510 At/m.
Find the current in the coil.(4.19 A)
Compute the values of permeability and relative permeability of the core.1.57 x 10‐3 Wb/Am, 1250 Wb/Am)
If the air‐gap is zero, find the current in the coil for the same flux density (0.8 T) in the core.(0.368 A)
Example 8
A magnetic circuit comprises three parts in series each having a uniform cross‐sectional area (A). They are:a) a length of 80 mm and A= 50 mm2
b) a length of 60 mm and A = 90 mm2
c) an air gap of length 0.5 mm and A = 150 mm2
A coil of 4000 turns is wound on part b) and the flux density in the air gap is 0.3 T. Assuming that all the flux passes through the given circuit, and the relative permeability is 1300, estimate the coil current to produce such a flux density
(45.43mA)
Series‐Parallel Magnetic Circuit(coil on left limb)
1ℜ
3S 2S
2233
3311
321
lHlH2looplHlHNI1loop
LawsKirchoff
=
+=
Φ+Φ=Φ
::
:
Series‐Parallel Magnetic Circuit(coil on centre limb)
2233
133
213
lHlHNI2looplHlHNI1loop
LawsKirchoff
+=
+=
Φ+Φ=Φ
::
:
`
1S
3S
2S
Series‐Parallel Magnetic Circuit(coil on centre limb, with air gap)
22ss33
11ss33
213
lHlHlHNI2looplHlHlHNI1loop
LawsKirchoff
++=
++=
Φ+Φ=Φ
::
:
1S 3S2S
gℜ
Hysteresis
The relationship between B and H is not linear for the types of iron used in motors and transformers.
dBHAlWW
B
v ∫==0
Hysteresis
Dr. Mohd Junaidi Abdul Aziz
Hysteresis
The relationship between B and H is complicated by non‐linearity and “hysteresis”
Can be used to calculate µ
Hysteresis
Hysterisis
Hysteresis Loss
Hysteresis loopUniform distribution
From Faraday's law
Where A is the cross section area
Dr. Mohd Junaidi Abdul Aziz
Field energyInput power :
Input energy from t1 to t2
where Vcore is the volume of the core
Hysteresis Loss
Dr. Mohd Junaidi Abdul Aziz
One cycle energy loss
Where ABH is the closed area of B‐H hysteresis loop
Hysteresis power loss
where f is the operating frequency and T is the period
Hysteresis Loss
Dr. Mohd Junaidi Abdul Aziz
Hysteresis Loss
Empirical equation
Summary : Hysteresis loss is proportional to f and ABH
Dr. Mohd Junaidi Abdul Aziz
Hysteresis Loss
Eddy Current Loss
Eddy currentAlong the closed path, apply Faraday's law
where A is the closed areaChanges in B → = BA changes
→induce e.m.f along the closed path→produce circulating circuit (eddy current) in the core
Eddy current losswhere R is the equivalent resistance along the
closed path
Calculation of eddy current lossFinite element analysisUse software: Ansys®, Maxwell®, Femlab®, etc
Eddy Current Loss
Dr. Mohd Junaidi Abdul Aziz
How to reduce Eddy current lossUse high resistively core material
e.g. silicon steel, ferrite core (semiconductor)
Use laminated coreTo decrease the area closed by closed path
Lamination thickness• 0.5~5mm for machines, transformers at line frequency• 0.01~0.5mm for high frequency devices
Eddy Current Loss
Core LossesHysterisis loss
the loss of power in the core due to the hysterisis effectProportional to frequency
Eddy current losspower loss occurs when the flux density changes rapidly in the coreProportional to the square of the frequency
.
losscurrenteddyPlosshysteresisPwhere
PPP
e
h
ehc
==
+=
Core Losses