EEB326 Electrical Machines I_ Lectures 1-3-2013

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  • 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

    law voltageKichholfs toanalogous is Which

    and

    T herefore 0 magnet icsin law sGauss' From

    Where

    11

    00

    sr

    n

    l

    l

    n

    k

    kk

    gc

    gc

    ggg

    g

    g

    c

    g

    ggccc

    cc

    cc

    c

    ccc

    ggcc

    nn

    FR

    RRF

    B.da

    RlA

    lB

    lHRlA

    lB

    lH

    lHlHNiF

    sr

  • 02

    0

    2

    0

    0

    2

    0

    2

    energy theincrease torequired force T he

    gap.air theof volumein the increase the torelated isenergy addit ional T he

    2

    energy addit ional T he

    done. work the todue is

    of gapair of volumeaddit ional in the storedenergy magnet ic addit ional T he

    .N force aby of distance aaway surfaces theof one movingConsider

    2

    gapair in storeddensity energy Magnet ic

    gg

    g

    g

    g

    g

    ABP

    dAB

    VWdPW

    dA

    Pd

    BW

  • 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 prod...