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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.html8/10/2019 dc motor 1.ppt
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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.html8/10/2019 dc motor 1.ppt
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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.html8/10/2019 dc motor 1.ppt
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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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