dc motor 1.ppt

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    Armature

    comutator

    Teras Kutub

    Belitan

    medanCarbon

    brush holder

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    A typical DC motor

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    (a) Armature and commutator segments. (b) Armature prior to the coil's wire

    being installed. (c) Coil of wire prior to being pressed into the armature. (d) A

    coil pressed into the armature. The end of each coil is attached to a

    commutator segment.

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    (a) This diagram shows the

    location of the pole pieces in

    the frame of a DC motor. (b)

    This diagram shows anindividual pole piece. You can

    see that it is made of

    laminated sections. The field

    coils are wound around the

    pole pieces.

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    The armature (rotor) of a DC motor has coils of wire wrapped around its

    core. The ends of each coil are terminated at commutator segments

    located on the left end of the shaft. The brushes make contact on the

    commutator to provide current for the armature.

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    The stationary part of a DC motor has the field coils mounted in it.

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    Belitan

    medan

    Belitan Kutub

    Antara

    Teras Kutub

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    Simple electrical diagram of DC shunt motor. This

    diagram shows the electrical relationship between the

    field coil and armature.

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    BELITAN GELOMBANG

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    Belitan ombak/gelombang C = 2

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    Belitan gelombong memerlukan :-

    - Sekurang-kurangnya dua berus tetapibilangan berus maksimum sama

    banyaknya dengan kutub .

    - Hanya dua laluan selari melalui angkerdalam satu belitan gelombang yang

    lengkap tanpa mengira bilangan berus .

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    BELITAN LAPIS, C = 2P

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    Secara am , belitan tindih/lapis

    memerlukan:-

    Berus sama banyak dengan kutub

    Laluan selari melalui angker samabanyak dengan kutub

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    MOTOR SIRI AT

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    Electrical diagram of series motor. Notice that the series field is

    identified as S1 and S2.

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    The relationship between series motor speed and the armature

    current.

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    MOTOR PIRAU AT

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    Diagram of DC shunt motor. Notice the shunt coil is identified as a coil of

    fine wire with many turns that is connected in parallel (shunt) with thearmature.

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    curve that shows the armature current versus the armature speed

    for a shunt motor. Notice that the speed of a shunt motor is nearly

    constant.

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    MOTOR MAJMUK AT

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    (a) Diagram of a cumulative compound motor, (b) Diagram of a differential

    compound motor, (c) Diagram of an interpole compound motor.

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    (a) Characteristic curve of armature current versus speed for the

    differential compound motor and cumulative compound motor, (b)

    Composite of the characteristic curves for all of the DC motors.

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    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/comtat.html
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    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/comtat.html
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    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/comtat.html
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    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motdc.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/comtat.html
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    Contoh Soalan:

    Sebuah motor at di sambung kpd bekalan

    240V. Rintangan angker 0.2. Tentukan

    nilai DGE balik jika arus angker ialah 50A

    Eb = VIaRa= 240(50 x 0.2)

    = 24010

    = 230V

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    The armature of dc machine has a resistance of 0.25

    and is connected to a 300V supply. Calculate the emf

    generated when it is running :a. as a generator giving 100A.

    b. as a motor taking 80A

    a. As a generator, generated emf:E = V + IaRa

    = 300 + (100 x 0.25) = 300 + 25= 325V

    b. As a motor, generated emf (or back emf):

    Eb = VIaRa

    = 300(80 x 0.25) = 30020 = 280V

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    Daya Kilas (Tork) Mesin AT

    Dari persamaan; V = Eb + IaRa

    Darabkan persamaan diatas dgn Ia:

    VIa = EbIa + IaRa

    VIa = Jumlah kuasa elektrik yg dibekal

    kpd angker

    EbIa= Jumlah kuasa mekanikal ygdihasilkan oleh angker.

    IaRa = Kehilangan kuasa oleh belitan

    angker

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    Jika T ialah tork dlm N-m,

    Oleh itu kuasa mekanikal yg terhasil adalah:

    T= 2nT = EbIa

    Tork, T = EbIa

    2

    n

    Juga : Eb = 2pZn

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    J g p

    c

    P = pasang kutub = fluks/kutub (Wb)

    Z = Bil. Pengalir angker

    n = kelajuan dlm rev/sc = 2 ; belitan ombak

    = 2p ; belitan lapis @ tindih

    Formula diatas boleh juga ditulis spt berikut:

    Eb = 2pZN

    60c

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    2nT = EbIa = ( 2pZn ) Ia

    c

    T = pZIa

    c

    T

    Ia

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    Sebuah motor dc 8 kutub dgn lilitan ombak mempunyai 900 pengalir

    angker. Fluks berguna per kutub ialah 25mWb. Tentukan nilai daya

    kilas apabila arus 30A mengalir melalui angker.

    T = pZIa

    c

    = 4 x 25 x 10-3x 900 x 30

    x 2= 429.7 Nm

    Tentukan nilai T yg terbina apabila motor at dgn voltan bekalan

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    Tentukan nilai T yg terbina apabila motor at dgn voltan bekalan

    350V mempunyai rintangan angker 0.5dan kelajuan putaran ialah

    15rev/s. Arus angker ialah 60A

    T = EbIa2n

    Eb = VIaRa

    = 350(60)(0.5)= 35030 = 320V

    T = EbIa = 320 x 60

    2n 2 x 3.14 x 15

    = 203.7 Nm

    A six pole lap wound motor is connected to a 250V dc

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    A six-pole lap-wound motor is connected to a 250V dc

    supply. The armature has 500 conductors and a resistance

    of 1 ohm. The flux per pole is 20mWb. Calculate (a) the

    speed and (b) the torque developed when the armaturecurrent is 40A

    a. Eb = VIaRa = 250(40 x 1) = 210V

    Eb = 2pZN60c

    N = 60cEb

    2pZ

    = 60 x 2 x3 x 2102 x 3 x 20 x 10-3x 500

    = 1260 rpm

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    b.T = EbIa

    2n

    = 210 x 40 x 60

    2 x x 1260

    = 63.7 Nm

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    DC machine losses

    The principal losses of machine are:a. Copper lossdue to I2R heat losses in

    the armature and field windings

    b. Iron (core) lossdue to hysteresis &eddy-current losses in the armature.

    This loss can be reduced by

    constructing the armature of silicon steellaminations.

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    c. Friction & windage lossesdue to bearing& brush contact friction & loss due to

    air resistance against moving parts(called windage)

    d. Brush contact lossbetween the brushes& commutator. This loss isapproximately propotional to the load

    current

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    Kecekapan motor AT

    Kecekapan = kuasa keluaran X 100%

    kuasa masukan

    Kuasa masukan = VI

    Kuasa keluaran = kuasa masukan - kehilangan

    kuasa

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    Kehilangan kuasa = kehilangan kuprum + kehilangan besi +

    geseran + angin

    = Ia Ra + If V + C

    C = kehilangan besi + geseran + angin

    Kecekapan = kuasa masukan - kehilangan kuasa X 100%

    kuasa masukan

    = 1 - kehilangan x 100%

    kuasa masukan

    Sebuah motor pirau 320V mengambil arus 80A dan berputar pd

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    kelajuan 1000 rpm. Jika jumlah kehilangan besi, geseran dan angin

    ialah 1.5KW. Diberi rintangan medan pirau Rf = 40, rintangan angker

    Ra = 0.2, tentukan kecekapan motor tersebut

    Arus medan, If = V/Rf = 320 / 40 = 8A Arus angker Ia = IIf = 808 = 72A

    Kuasa masukan = VI = 320 x 80 = 25600 W

    Kehilangan kuasa =Ia Ra + If V + C

    = (72) x 0.2 + 8 x 320 + 1500

    = 1036.8 + 2560 +1500 = 5096.8 W

    Kecekapan = 1 - kehilangan x 100%

    kuasa masukan

    = 1 - 5096.8 x 100%

    25600

    = (10.1991) x 100%

    = 80.1%

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