EEB326 Electrical Machines I
Lectures Weeks 1-3
Magnetic Circuits
Magnetic fields
a) Unlike poles attract b) Like poles repel
Terms and symbolsMagnetic flux (crossing a surface of area A m2)
[Weber , abbreviated Wb]
Magnetomotive force
Fm = NI (Amperes ) where N is number of turns and I is
current.
Magnetic field strength or magnetising force
length. is current, is turns,ofnumber is Where lINl
NIH
Example 1: An
electromagnet of
square cross section
produces a flux
density of 0.45 T
when the flux is 720
Wb. Determine the dimensions of the cross section.
l= mean length
N
I
d
w
m 04.0
m 45.0
10*720 26
2
d
Bd
Example 2:
A coil of 750 turns is uniformly wound on a circular magnetic
circuit of mean diameter 30 cm. Find the current in the coil
when the magnetising force is 8000 A/m.
Solution
A 10.0531
A 750
10*30**8000 2N
HlI
Magnetic Flux Density
B=H [Tesla (T)] where is permeability of the material
=r0 where r is relative permeability of the material, and
0 = 4 x 10-7 H/m is permeability of free space.
Notes: 1T = 1 Wb/m2.
B=/A
Example 3: A magnetic pole face has a rectangular section of 300 mm by 150 mm. Calculate the flux density when the total flux from the pole is 300 Wb.
Solution:
B = (300*10-6 Wb)/(300*10-3*150*10-3 m2)=0.0066667 T
=6.6667 mT
in vacuumdensity flux
materialin density flux r
Example 4:
An iron ring of mean diameter 10 cm is uniformly wound with 2000
turns of wire. When a current of 0.25A is passed through the coil a flux
density of 0.4 T is set up in the iron. Find (a) the magnetising force and
(b) the relative permeability of the iron under these conditions.
a.l = d = 10 cm = 10 10-2 m,N = 2000 turns, I = 0.25A and B = 0.4 T
b. B = 0rH, hence
Reluctance
F =NI= Magneto Motive Force or
MMF = # of turns * Current passing through it
NIB
lor NIA
lor
or)/( A
NI
l
NIor
F = NI = Hl (why!)
A
l
= Reluctance of magnetic path
Example 5:
Find (a) The reluctance of a piece of mumetal of length
150 mm and cross-sectional area 1800 mm2 when the
relative permeability is 4000. (b) The absolute permeability
of the mumetal.
= 0r = (4 10-7)(4000)
= 5.027 10-3 H/m
Example 6:
A mild steel ring has a radius of 50 mm and a cross-sectional
area of 400 mm2. A current of 0.5A flows in a coil wound
uniformly around the ring and the flux produced is 0.1 mWb. If
the relative permeability at this value of current is 200 find (a)
the reluctance of the mild steel and (b) the number of turns on
the coil.
l = 2r = 2 50 10-3 m, A = 400 10-6 m2,I = 0.5A, = 0.1 10-3 Wb and r = 200
from whichm.m.f. = S i.e. NI = S Hence, number of turns
Composite series magnetic circuits
For a series magnetic circuit having n parts the total
reluctance is the sum of the individual reluctances.
This is similar to the series connection of resistances
Example 7:
A closed magnetic circuit of cast steel contains a 6 cm
long path of cross-sectional area 1 cm2 and a 2 cm path
of cross-sectional area 0.5 cm2. A coil of 200 turns is
wound around the 6 cm length of the circuit and a
current of 0.4A flows. Determine the flux density in the 2
cm path, if the relative permeability of the cast steel is
750.
Total circuit reluctance S = S1 + S2= (6.366 + 4.244) 105
= 10.61 105/H
Flux density in the 2 cm path,
Electric and magnetic circuit similarities
Electric and magnetic quantities correspondences
Electric quantity Magnetic quantity
Current I Magnetic flux
Current density J Magnetic flux density B
Conductivity Permeability
Electromotive force =
resistance x I
Magnetomotive force = reluctance x
Electric field intensity E Magnetic field intensity H
Conductance = 1/resistance Permeance = 1/reluctance
Resistance =l/A Reluctance =l/A
Magnetic circuit laws
Consider a magnetic circuit with an air gap. Note that the fringing at
the air gap increases the effective cross sectional area.
lyrespect ive sources
and sreluctance of numbers are and Where
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and
T herefore 0 magnet icsin law sGauss' From
Where
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Tutorial: Lifting electromagnet design.
The yoke of a lifting electromagnet is made up of 20
laminations each 1.5 mm thick. Each lamination is cut
from a 80 mm x 35 mm rectangular plate. A portion 40
mm x 15 mm is cut out of the plate so that each pole face
is 20 mm wide. The coil of N turns is wound on the yoke.
The supply to the coil will be 0.5 Amps DC. The
electromagnet is to lift a mild steel block of mass 0.5 kg
and dimensions of 100 mm length, 20 mm width and 30
mm depth. The air gap can have a maximum value of 1
mm.
Determine the number of turns when the air gap is 0.1
mm. Use the BH curves provided.
Lifting electromagnet design
.element of area sect ional cross is Where
element each in density flux Calculate 5
circuitin flux magnet ic Calculate 4
area gapair is Where2
densityflux desired Calculate 3
. force lift ing desired thehence and lifted be tomass Determine 2
net .electromag lift ing of dimensions Determine 1
proceduredesign net electromag Lift ing
0
jAA
B
jB
*AB
AA
FB
F
j
j
circuitj
j
grequiredcircuit
g
g
desireddesired
desired
available
m total
j
jcircuitm total
jj
j
j
jj
I
FN
R F
A
lR
H
turnsofnumber determinecurrent available From 9
force ivemagnetomot Calculate 8
elementcircuit each for reluctance Calculate 7
curves BH smaterial' from and Find 6
Consider a magnetic circuit with a T joint
analysiscircuit magnet icin applicable are
analysis nodal andmesh assuch s, techniqueanalysiscircuit DC
junct ion aat branches ofnumber theis Where
0
generalin Or
0
joint T the tolaw Gauss Applying
1
3
1
n
n
k
k
k
k
Magnetic circuit
model of
Permanent
Magnet.
Characteristics of
permanent
magnets are in
the second
quadrant.
mmmmcm
mm
m
mc
m
m
mcmmm
m
cmmcm
c
rm
m
FRlHA
l
lH
BlHHlH
HHHHH
BB
B-Hl
A
magnet theacross drop" voltage" magnet ic T he
.nearly ismagnet permanent theofty permeabili the,
formula thehas curveat ion demagnet iz T he
st ic.characteri linear a and length
, area sect ional cross ofmagnet permanent aConsider
0
m
Model of a Permanent Magnet of cross sectional area Am, length lm and a linear B-H characteristic . Note that Hm is in opposite direction to Bm.
.ofdirection opposite in the being negative is Where
isty permeabili magnetic thecurve,ation demagnetiznonlinear aFor
mm
cm
mm
BH
HH
B
Permeability
Where r typically is 1000-100000in magnetic cores
= r 0
Hysteresis loop
0ab: magnetising curve, from
demagnetised state.
by: Saturation flux density
0c: remanent flux density or
remanence, H reduced to zero.
0d: Coercive force required to reduce flux to zero.
bcde: curve formed by reducing H.
ex: Saturation flux density.
efgb: Curve formed by increasing H.
Hysteresis: Changes in B lag behind
changes in H.
bcdefgb: Hysteresis loop
Hysteresis lossEnergy is expended in ferromagnetic
materials to change the alignment of
domains during cycles of magnetisation.
The energy appears as heat and is
called hysteresis loss.
Hysteresis loss is proportional to the
area of the hysteresis loop.
The area of the hysteresis loop varies
with the hardness/softness of materials.
Figure a: hard material, high remanence,
large coercivity
Figure b: soft material large remanence,
small coercivity
Figure c: Ferrite small hysteresis
a) Unmagnetised domains
b) Magnetised domains
Magnetisation curve
Material Relative Permeability
Mild steel 200 800
Silicon iron 1000 - 5000
Cast steel 300 - 900
Numetal 200 000
Stalloy 500 - 6000
Electromagnetism
Magnetic field due to an electric current
Direction of magnetic field inside coil given by either
screw rule or grip rule
Electromagnets
Electric bell -single stroke.
When the push button is pressed a current flows and the armature is attracted to the electromagnet. The striker on the armature hits the gong. When the circuit is broken the electromagnet demagnetises and the armature is pulled back by the spring.
Relay
Contacts are
closed, or
opened,
instead of
striking a bell.
The contacts are
for another
electrical circit.
Lifting magnet
Coil C is wound around
cast iron core P. A
protective non-magnet
covering R is over the
face of the
electromagnet. The
magnetic paths M pass
through the magnetic
material to be lifted.
Telephone
Receiver
Converts electrical
signals to sound
waves. The
diaphragm
vibrates
according to the
electrical signals
transmitted to the
receiver.
Force on a current carrying conductor
F=BIl sin
Where B is magnetic flux density, I is current, l is length
of conductor and is angle between conductor and magnetic field.
Loudspeaker
When an electric current flows in the coil it produces a force that moves the cone backwards and forwards according to the direction of the current. The cone transfers the force to the air, producing sound waves.
A current carrying
conductor in a
magnetic circuit
experiences a force
due to the
interaction of the
two magnetic
fluxes; the one for
the magnetic field
and that due to the
current.
Flemings left hand rule
Let the thumb, first finger
and second finger of the
left hand be extended
such that they are all at
right-angles to each other,
If the first finger points in
the direction of the
magnetic field, the second
finger points in the
direction of the current,
then the thumb will point
in the direction of the
motion of the conductor
Simple dc motor
Magnetic field supplied by permanent magnet.
Current supplied to coil through brushes and commutator bar.
Commutator allows for change of direction of current.
Coil rotates due to the forces acting on it.
Brushes provide contact to a moving commutator.
A practical motor has more than one winding
Example 8:
A coil is wound on a rectangular former of width 24 mm and
length 30 mm. The former is pivoted about an axis passing
through the middle of the two shorter sides and is placed in a
uniform magnetic field of flux density 0.8 T, the axis being
perpendicular to the field. If the coil carries a current of 50
mA,
Determine the force on each coil side
(a) for a single-turn coil,
(b) for a coil wound with 300 turns.
Solution
a Flux density B = 0.8 T, length of conductor lying at right-angles
to field l = 30 mm = 30 10-3 m and current I = 50 mA = 50
10-3 A.
For a single-turn coil, force on each coil side
F = BIl = 0.8 50 10-3 30 10-3
= 1.2 10-3 N or 0.0012 N
b When there are 300 turns on the coil there are effectively 300
parallel conductors each carrying a current of 50 mA. Thus the
total force produced by the current is 300 times that for a
single-turn coil. Hence force on coil side, F = 300 BIl = 300
0.0012 = 0.36 N
Elementary dc motor summary
Back emf: a voltage is induced in a motor coil as it revolves
Note that a dc motor can also operate as a dc generator
DC motor equivalent circuit: Note the two flux systems
E0 = K where =2n/60, n is speed in rpm (revolutions per minute)
The back emf Eb = E0 = Km (1) Where m is mechanical speed of rotation =2n/60, n=speed in rpm (revolutions per minute), K is a factor that depends on coil geometry.
The armature current Ia =(VT - E0)/Ra (2) Note that the starting current is very high since E0 is zero at start.
The torque developed T = KIa (3) Substituting for Iafrom (2 ) T = K(VT-E0)/Ra = (K/Ra)(VT - Km)
The power developed by the motor Pdev=E0Ia = KmIa(=Mechanical power =mT= KmIa)
Torque speed characteristic
Operating point depends on load characteristic
Speed regulation: One of the measures of performance
load fullat speed is
load noat speed is Where
100
motor. the toapplied is load full as speed
in change theas defined is regulat ion Speed
full load
no load
full load
full loadno load
N
N
%xN
- NNlationSpeed regu
DC machine classifications Depends on arrangement
of supplies to the field and armature windings.
a. Separately excited machine when the field winding has a
separate supply
b. Self excited machine when field winding is connected to
the main motor supply. Three types of connections: -
i. Shunt connected machine when the field is across the main supply
ii. Series connected machine when the field and armature windings
are in series.
iii. Compound connected machine when both shunt and series
connections are used. The connections can be cumulative
compounding (fluxes additive) or differential compounding (fluxes
in opposition)
Separately
excited motor
m
a
dev KVTR
KT
Shunt excited motor, has
same torque speed
characteristic as a
separately excited motor
m
a
dev KVTR
KT
Series excited motor
2
2
mffa
Tf
devKKRR
VKKT
Derivation of torque speed characteristic for series motor
2
2
2
2
2
a
'
'
(3)in for ngSubst itut i
circuit armature thefrom found is
(3) ...
(2) ...
and (1) ...
mfa
T
mffa
Tf
dev
a
mffa
Ta
mffaamfaabfaaT
a
fdev
af
affadev
KRR
VK
KKRR
VKKT
I
KKRR
VI
KKRRIKRRIERRIV
I
IKKT
IK
IIIIKT
Example
A 230 V 10 hp DC shunt motor delivers power to a load at
1200 rpm. The armature current drawn by the motor is
200 A. The armature circuit resistance of the motor is 0.2
and the field resistance is 115 .
a) If the rotational losses are 500 W, what is the value of the load
torque.
b) What is the starting current of the motor
Solution part (a)
VT = 230 V
Ra = 0.2 Ia = 200 A
Prot = 500 W
N = 1200 rpm
Back emf induced in the armature
Eb = VT - IaRa = 190 V
Power developed in the rotor
Pdev = EbIa = 38000 W
Power delivered to the load
Pload =Pdev - Prot = Pout = 37500 W
Load torque = Pload/m m = 2N/60 = 125.6637 rads/sTload = 298.4155 N m
Solution part (b)
Starting current Ia start = VT / Ra = 230/0.2 A = 1150 A
Faradays law of electromagnetic induction
Faradays law
If the flux linking a loop (or turn) varies as a function of
time, a voltage is induced between its terminals.
The value of the induced voltage is proportional to the rate
of change of flux.
s changesflux which theduring interval t imeist
Wb coil theinsideflux of change is
coil, of turnsofnumber is N
,V voltageinduced is E
- :Where
t
NE
When the magnetic field in
a coil changes a voltage is
induced in the coil. This
voltage will drive a current
in a circuit to which the
coil is connected.
Electromagnetic Induction
When a conductor is moved across a magnetic field or
vice versa so as to cut through the lines of force (or flux),
an electromotive force (e.m.f.) is produced in the
conductor.
If the conductor forms part of a closed circuit then the
e.m.f. produced causes an electric current to flow round
the circuit.
Hence an e.m.f. (and thus current) is 'induced' in the
conductor as a result of its relative movement across the
magnetic field.
This effect is known as 'electromagnetic induction'
Lenz's law states:
The direction of an induced e.m.f. is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.
An alternative method to Lenz's law of determining relative directions is given by Fleming's Right-hand rule (often called the geneRator rule) which states:
Let the thumb, first finger and second finger of the right hand be extended such that they are all at right angles to each other . If the first finger points in the direction of the magnetic field and thethumb points in the direction of motion of the conductor relative to the magnetic field, then the second finger will point in the direction of the induced e.m.f.
Conductor moving in a magnetic field
If a wire is moved
through a
magnetic field then
a voltage can be
induced across the
ends the wire.
Induced voltage
E = Blv sin volts
Where: -
B Magnetic flux density [Wb/m2]
l length of conductor
v velocity of conductor [m/s]
angle between conductor and magnetic field.
Example 9:
A conductor 300 mm long moves at a uniform speed of 4 m/s at
right-angles to a uniform magnetic field of flux density 1.25 T.
Determine the current flowing in the conductor when (a) its
ends are open-circuited, (b) its ends are connected to a load of
20 resistance.
Solution:
When a conductor moves in a magnetic field it will have an e.m.f.
induced in it but this e.m.f. can only produce a current if there
is a closed circuit.
Induced e.m.f.
a. If the ends of the conductor are open circuited no current
will flow even though 1.5V has been induced.
b. From Ohm's law,
Figure shows a view of a looped conductor whose sides are
moving across a magnetic field.
The left-hand side is moving in an upward direction (check using
Fleming's right-hand rule), with length l cutting the lines of flux
which are travelling from left to right. By definition, the induced
e.m.f. will be equal to Blv sin and flowing into the page.
The right-hand side is moving in a downward direction (again,
check using Fleming's right-hand rule), with length l cutting the
same lines of flux as above. The induced e.m.f. will also be equal
to Blv sin but flowing out of the page.
Therefore the total e.m.f. for the loop conductor = 2Blv sin
Now consider a coil made up of a number of turns N . The total
e.m.f. E for the loop conductor is now given by:
E = 2NBlv sin
Example 10:
A rectangular coil of sides 12 cm and 8 cm is rotated in a
magnetic field of flux density 1.4 T, the longer side of the coil
actually cutting this flux. The coil is made up of 80 turns and
rotates at 1200 rev/min. (a) Calculate the maximum generated
e.m.f. (b) If the coil generates 90 V, at what speed will the coil
rotate?
Solution
a Generated e.m.f. E = 2NBLv sin
where number of turns, N = 80,
flux density, B = 1.4 T,
length of conductor in magnetic field, l = 12 cm = 0.12 m
and for maximum e.m.f. induced, = 90, from which, sin = 1
Hence, maximum e.m.f. induced,
E = 2NBlv sin
= 2 80 1.4 0.12 1.6 1
= 135.1 volts
b Since E = 2NBlv sin
then 90 = 2 80 1.4 0.12 v 1
An alternative method of determining (b) is by direct
proportion.
Since E = 2NBlv sin , then with N, B, l and being constant, E v
If from (a), 135.1 V is produced by a speed of 1200 rev/min,
then 1V would be produced by a speed of
Hence, 90V would be produced by a speed of
90 8.88 = 799 rev/min
Generators
When the coil is
horizontal, the
induced current is
maximum, as the coil
is cutting across the
field lines at right
angles as it moves.
When the coil is
vertical, the induced
current is zero, as
the coil is moving
parallel to the
magnetic field lines
If we start timing from when the coil is vertical, then
at t=0, = BxA
N S
B
A
If the coil rotates with speed , then after time t the coil will have turned through angle = t
The flux will now be = Bcost x A
N S
A
B
Faradays Law says:
For a coil of N turns:
tBAV
dt
tBAdV
dt
dV
sin
)cos(
tBANV sin
The formula for alternating generator voltage is
often written as:
Where Vmax=BAN
This produces a voltage-time graph that looks like a
sine curve
NB. Similarities to SHM!!
tVV sinmax
To generate D.C, split rings are used.
To generate A.C, slip rings are used
DC generator equivalent circuit
Separately excited
Ia = I = load current
Load voltage VL = VT = Eg IaRa VbrWhere
Eg is generated voltage,
VT is terminal voltage,
Ra is armature resistance,
Vbr is voltage drop across the brushes (may be assumed
negligibly small)
Power input (from prime mover) Pin = Pm = EgIa
Power output = VL Ia =
Magnetisation curve
Voltage current characteristic for separately excited
generator
Self excited: Need for residual magnetism and forward
direction of rotation
Series connected generator equivalent circuit
Ia = If = IL = load current
Load voltage VL = VT = Eg Ia(Ra+Rf) VbrWhere
Eg is generated voltage,
VT is terminal voltage,
Ra is armature resistance,
Rf is field resistance
Vbr is voltage drop across the brushes (may be assumed
negligibly small)
Power input (from prime mover) Pin = Pm = EgIa
Power output = VL Ia =
Shunt connected generator equivalent circuit
Ia = If + ILWhere If = VL/Rf = field current where Rf is field resistance
IL= load current
Generated voltage Eg= VL + IaRa + VbrWhere
Eg is generated voltage,
VL is terminal / oad voltage,
Ra is armature resistance,
Vbr is voltage drop across the brushes (may be assumed
negligibly small)
Power input (from prime mover) Pin = Pm = EgIa
Power output = VL Ia =
Shunt connected
generator
OA, OB, OC slopes
give critical
resistances at the
indicated speeds.
Rf should be less
than this value for
the particular
speed.
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