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.

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and

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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

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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

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n

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k

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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

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.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.