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DC MACHINES
Dr. Abdulr-Razaq SH. Hadde
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DC Motor
• The direct current (dc) machine can be used as a motor or as a generator.
• DC Machine is most often used for a motor. • The major advantages of dc machines are the
easy speed and torque regulation. • However, their application is limited to mills,
mines and trains. As examples, trolleys and underground subway cars may use dc motors.
• In the past, automobiles were equipped with dc dynamos to charge their batteries.
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DC Motor
• Even today the starter is a series dc motor • However, the recent development of power
electronics has reduced the use of dc motors and generators.
• The electronically controlled ac drives are gradually replacing the dc motor drives in factories.
• Nevertheless, a large number of dc motors are still used by industry and several thousand are sold annually.
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Construction
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DC Machine Construction
Figure 8.1 General arrangement of a dc machine
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DC Machines
• The stator of the dc motor has poles, which are excited by dc current to produce magnetic fields.
• In the neutral zone, in the middle between the poles, commutating poles are placed to reduce sparking of the commutator. The commutating poles are supplied by dc current.
• Compensating windings are mounted on the main poles. These short-circuited windings damp rotor oscillations. .
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DC Machines
• The poles are mounted on an iron core that provides a closed magnetic circuit.
• The motor housing supports the iron core, the brushes and the bearings.
• The rotor has a ring-shaped laminated iron core with slots.
• Coils with several turns are placed in the slots. The distance between the two legs of the coil is about 180 electric degrees.
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DC Machines
• The coils are connected in series through the commutator segments.
• The ends of each coil are connected to a commutator segment.
• The commutator consists of insulated copper segments mounted on an insulated tube.
• Two brushes are pressed to the commutator to permit current flow.
• The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.
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DC Machines
• The rotor has a ring-shaped laminated iron core with slots.
• The commutator consists of insulated copper segments mounted on an insulated tube.
• Two brushes are pressed to the commutator to permit current flow.
• The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.
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DC Machines
• The commutator switches the current from one rotor coil to the adjacent coil,
• The switching requires the interruption of the coil current.
• The sudden interruption of an inductive current generates high voltages .
• The high voltage produces flashover and arcing between the commutator segment and the brush.
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DC Machine Construction
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Shaft
Brush
Coppersegment
Insulation
RotorWinding
N S
Ir_dcIr_dc/2
Rotation
Ir_dc/2
Ir_dc
12
3
4
5
6
7
8
Polewinding
Figure 8.2 Commutator with the rotor coils connections.
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DC Motor Operation
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DC Motor Operation
• In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field.
• The rotor is supplied by dc current through the brushes, commutator and coils.
• The interaction of the magnetic field and rotor current generates a force that drives the motor
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Shaft
Brush
Coppersegment
Insulation
RotorWinding
N S
Ir_dcIr_dc/2
Rotation
Ir_dc/2
Ir_dc
12
3
4
5
6
7
8
Polewinding
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DC Motor Operation
• Before reaching the neutral zone, the current enters in segment 1 and exits from segment 2,
• Therefore, current enters the coil end at slot a and exits from slot b during this stage.
• After passing the neutral zone, the current enters segment 2 and exits from segment 1,
• This reverses the current direction through the rotor coil, when the coil passes the neutral zone.
• The result of this current reversal is the maintenance of the rotation.
(a) Rotor current flow from segment 1 to 2 (slot a to b)
Vdc30
NS
Bv
v
a
b
1
2
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
30NS Vdc
a
b
1
2
B
v v
Ir_dc
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DC Generator Operation
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DC Generator Operation
• The N-S poles produce a dc magnetic field and the rotor coil turns in this field.
• A turbine or other machine drives the rotor.
• The conductors in the slots cut the magnetic flux lines, which induce voltage in the rotor coils.
• The coil has two sides: one is placed in slot a, the other in slot b.
30NS Vdc
Bv
v
a
b
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
30NS Vdc
a
b
1
2vv
B
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
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DC Generator Operation
• In Figure 8.11A, the conductors in slot a are cutting the field lines entering into the rotor from the north pole,
• The conductors in slot b are cutting the field lines exiting from the rotor to the south pole.
• The cutting of the field lines generates voltage in the conductors.
• The voltages generated in the two sides of the coil are added.
30NS Vdc
Bv
v
a
b
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
30NS Vdc
a
b
1
2vv
B
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
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DC Generator Operation
• The induced voltage is connected to the generator terminals through the commutator and brushes.
• In Figure 8.11A, the induced voltage in b is positive, and in a is negative.
• The positive terminal is connected to commutator segment 2 and to the conductors in slot b.
• The negative terminal is connected to segment 1 and to the conductors in slot a.
30NS Vdc
Bv
v
a
b
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
30NS Vdc
a
b
1
2vv
B
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
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DC Generator Operation
• When the coil passes the neutral zone: – Conductors in slot a are
then moving toward the south pole and cut flux lines exiting from the rotor
– Conductors in slot b cut the flux lines entering the in slot b.
• This changes the polarity of the induced voltage in the coil.
• The voltage induced in a is now positive, and in b is negative.
30NS Vdc
Bv
v
a
b
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
30NS Vdc
a
b
1
2vv
B
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
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DC Generator Operation
• The simultaneously the commutator reverses its terminals, which assures that the output voltage (Vdc) polarity is unchanged.
• In Figure 8.11B – the positive terminal is
connected to commutator segment 1 and to the conductors in slot a.
– The negative terminal is connected to segment 2 and to the conductors in slot b.
30NS Vdc
Bv
v
a
b
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
30NS Vdc
a
b
1
2vv
B
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
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Generator
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DC Generator Equivalent circuit
• The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.
• This induced voltage is represented by a voltage source.
• The stator coil has resistance, which is connected in series.
• The pole flux is produced by the DC excitation/field current, which is magnetically coupled to the rotor
• The field circuit has resistance and a source
• The voltage drop on the brushes represented by a battery
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DC Generator Equivalent circuit
• Figure 8.12 Equivalent circuit of a separately excited dc generator.
RfRa
Vbrush
VdcEagVf
max
IfIag
Load
Mechanicalpower in
Electricalpower out
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DC Generator Equivalent circuit
• The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.
• The dc field current of the poles generates a magnetic flux
• The flux is proportional with the field current if the iron core is not saturated:
fag IK1
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DC Generator Equivalent circuit
• The rotor conductors cut the field lines that generate voltage in the coils.
• The motor speed and flux equations are :
vBNE grag 2
2gDv ggag DB
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DC Generator Equivalent circuit
• The combination of the three equation results the induced voltage equation:
• The equation is simplified.
agrggrg
grgrag NDBND
BNvBNE
222
fmfragrag IKIKNNE 1
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DC Generator Equivalent circuit
• When the generator is loaded, the load current produces a voltage drop on the rotor winding resistance.
• In addition, there is a more or less constant 1–3 V voltage drop on the brushes.
• These two voltage drops reduce the terminal voltage of the generator. The terminal voltage is;
brushaagdcag VRIVE
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Motor
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DC Motor Equivalent circuit
• Figure 8.13 Equivalent circuit of a separately excited dc motor• Equivalent circuit is similar to the generator only the current directions are
different
RfRa
Vbrush
VdcEamVf
max
IfIam
Mechanicalpower out
Electricalpower in
DC Powersupply
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DC Motor Equivalent circuit• The operation equations are:• Armature voltage equation
brushaamamdc VRIEV
The induced voltage and motor speed vs angular frequency
fmam IKE mn 2
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DC Motor Equivalent circuit• The operation equations are:• The combination of the equations results in
The current is calculated from this equation. The output power and torque are:
mamdcamfm RIVEIK
amamout IEP fammout IIKP
T
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Ideal Transformers
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I1 I2 I1 I2
V1 V2
V1 V2N1 N2
IT+ +- -
I1 I2
Z2V1 V2
N1 N2
The ideal transformer: V1/N1 = V2/N2 N1I1 = N2I2 S1 = V1I1* S2 = V2I2* S1 = S2 k = V2/V1 = N2/N1 V1 = V2/k I1 = k* I2 Z2 = V2/I2 Z2’ = V1/I1 = (V2/k) / (I2*k) = Z2/k2
Real transformer:
I2I1
N2N1
R1jX1 R2 jX2
LoadV1 V2Ic
R0 jX0
Im
V1
I2*k
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N2N1 IT
V2
+
-
V2V1
V1
V2 =
N1
N2k =
V1
I1 I2R1 jX1
Ic
Rc jXm
R2 jX2
Load
Im
N1 N2
I1 R1 jX1
Ic
Rc jXm
R2’ jX2’
Im
N1 N2
V1
I2
I2*k
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I1 I2 I1 I2
V1 V2
V1 V2N1 N2
IT+ +- -
I2I1
N2N1
R1jX1 R2 jX2
LoadV1 V2Ic
Rc jXm
Im
V1
I2*k
I1 R1 jX1
Ic
Rc jXm
R2’ jX2’
Im
N1 N2
V1
I2
Load
R2 & X2 are referred to winding 1
1
2
1
2
N
N
V
Vk
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I1 Req jXeq I2
V1
N1 N2
Load
Req = R1 + R2’
Xeq = X1 + X2’Neglecting excitation current
N1 N2
I2jXeqI1
V1 Load
V2
V2
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Determination of Req and Xeq from short-circuit test. Vsc is normally equal to five percent of the rated voltage of winding #1, or: Vsc = .05 x V1rated Measure Isc, Vsc, Psc Psc = /I2sc/* Rs1 Rs1 = Psc / (/I2sc/) Zsc1 = /Vsc1///Isc1/= /Req + jXeq/ = /Rs1 + jXs1/ Xs1 = Zsc2 - Rs2 Vsc = .05 x V2rated Measure Isc, Psc, Vsc The same procedure as before however, the short-circuit impedance is referred to
winding #2.
Real Transformer Ideal Transformer
Req jXeq
Vsc
Isc
Vsc
jXeqReqIsc
VscVsc
Isc
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The per-unit system: pu = Actual Quantity / Base Value of Quantity Name plate rating: An electric element has a nameplate rating which specifies the safe operating voltage, current and power for the element. Example: A light bulb is rated at 120V and 50Watts. Base Values: The base values are normally selected equal to base values. Pb = Qb = Sb = S1(rated)
Vb = VL-N(rated) Zb = Vb/Ib = Vb / (Sb/Vb) = Vb2/Sb Yb = 1/Zb Ppu = PA/Pb Qpu = QA/Qb Spu = SA/Sb Vpu = VA/Vb Ipu = IA/Ib Zpu = ZA/Zb
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The nameplate rating for transformers, generators and machines are given in percent (which is the same as pu, i.e. percent% = pu) using the device power rating as Sb and voltage rating as Vb. Example: A 3 generator is rated 300 MVA, 20 kV and ten percent reactance. Compute the impedance of the machine in ohms. Sb = 300 MVA/3 = 100 x 106 MVA Vb = 20 kV/ 3 = 11.56 x 103 V Zb = Vb2/Sb = (11.56 x 103)2 / 100 x 106 = 1.34 Zgen = j.1 x (1.34) = j.134
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Rs’ jXs’
V1(rated) V2(rated)
Rs jXs
V1(rated) V2(rated)
N1 N2 N1 N2
I1I1 I2
S(rated) S(rated)LV HV
a = N1
N2=
V1(rated)
V2(rated)
P.U. Equivalent Circuit of a Transformer. Select: Vb1 = V1(rated) Sb = S(rated) Select: Vb2/Vb1 = V2(rated) / V1(rated) or Vb2 = (V2(rated) / V1(rated))Vb1 Zb1 = Vb1/Ib1 = Vb1
2/Sb Sb = Vb1Ib1 Zb2 = Vb2/Ib2 = Vb2
2/Sb Ib2 = Sb/ Vb2 Zb1 / Zb2 = Vb1
2/ Vb22 = Vb1
2/( V2(rated) / V1(rated))2 Vb12 = 1/(1/a)2 = a2
Zb1 = Zb2 / k2
Zpu1 = Rs+jXs/Zb1 = Rs+jXs/ (Zb2/k2 ) = (Rs*k2) +(jXs*k2)/Zb2 = R’s+jX’s/ Zb2 = (Rs*k2 + jXs*k2)1/ Zb2 or Zpu1 = Zpu2
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Rs jXs
V1(rated) V2(rated)
N1 N2
Vb1 = V1(rated) Sb = S(rated 1 phase) Vb2 = V2(rated)
Zb1 = (Vb1)2/Sb Zb2 = (Vb2)2/Sb
Zpu
Vb1 Vb2
Zpu = Rs+jXs/Zb1
Vb1 Vb2
I1 pu
V1 pu
Rs/a2 + jXs/a2
V2(rated)
N1 N2
Sb = S1(rated) Vb2 = V2(rated)
Zb1 = (Vb1)2/Sb Zb2 = (Vb2)2/Sb
Zb2 = Zb1*k2
Zpu
Vb1
Vb2
Sb
Vb1
I1 pu
V1 pu V2 pu
Vb2
I2 pu
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Transformation between two bases:
Selection 1: Sb1 = SA Vb1 = VA
Then:Zb1 = (Vb1)2 / Sb1 Zpu1 = ZL/ Zb1
Selection 2: Sb2 = SB Vb2 = VB
Then:Zb2 = (Vb2)2 / Sb2 Zpu2 = ZL/ Zb2
Zpu2 / Zpu1 = ZL / Zb2 x Zb1 / ZL = Zb1 / Zb2
= (Vb1)2 / Sb1 x Sb2 / (Vb2)2
Zpu2 = Zpu1 (Vb1 / Vb2)2 (Sb2 / Sb1)
“1” = old“2” = new
Zpu (new) = Zpu (old) (Vb (old) / Vb (new))2 (Sb (new) / Sb (old))
Transformation between two bases:
Selection 1: Sb1 = SA Vb1 = VA
Then:Zb1 = (Vb1)2 / Sb1 Zpu1 = ZL/ Zb1
Selection 2: Sb2 = SB Vb2 = VB
Then:Zb2 = (Vb2)2 / Sb2 Zpu2 = ZL/ Zb2
Zpu2 / Zpu1 = ZL / Zb2 x Zb1 / ZL = Zb1 / Zb2
= (Vb1)2 / Sb1 x Sb2 / (Vb2)2
Zpu2 = Zpu1 (Vb1 / Vb2)2 (Sb2 / Sb1)
“1” = old“2” = new
Zpu (new) = Zpu (old) (Vb (old) / Vb (new))2 (Sb (new) / Sb (old))
ZL