Download Lecture Notes EEE 360

DC MACHINES
Dr. Abdulr-Razaq SH. Hadde
1
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.
2
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.
3
Construction
4
DC Machine Construction
Figure 8.1 General arrangement of a dc machine
5
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. .
6
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.
7
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.
8
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.
9
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.
10
DC Machine Construction
Rotation
Ir_dc/2
Brush
Ir_dc/2
Ir_dc
Shaft
Pole
winding
|
1
2
8
N
3
7
6
S
4
5
Insulation
Rotor
Winding
Ir_dc
Copper
segment
Figure 8.2 Commutator with the rotor coils connections.
11
DC Motor Operation
12
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
Rotation
Ir_dc/2
Brush
Ir_dc/2
Ir_dc
Shaft
Pole
winding
|
1
2
8
N
3
7
6
S
4
5
Insulation
Rotor
Winding
Ir_dc
Copper
segment
13
DC Motor Operation
•
N
30
Vdc
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
a
S
2
•
S
B
a
2
•
v
1
•
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.
30
v
v
N
Vdc
1
•
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
14
DC Generator
Operation
15
DC Generator Operation
v
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
• 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.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
16
DC Generator Operation
v
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
• 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.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
17
DC Generator Operation
v
B
a
S
N
1
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
• 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.
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
18
DC Generator Operation
• When the coil passes the
neutral zone:
S
N
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
• This changes the polarity
of the induced voltage in
the coil.
• The voltage induced in a
is now positive, and in b is
negative.
B
a
1
– 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.
v
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
19
DC Generator Operation
B
a
S
N
30
Vdc
2
b
v
Ir_dc
(a) Rotor current flow from segment 1 to 2 (slot a to b)
B
S
2
a
30
v
v
N
Vdc
1
– 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.
v
1
• The simultaneously the
commutator reverses its
terminals, which assures
that the output voltage
(Vdc) polarity is
unchanged.
• In Figure 8.11B
b
Ir_dc
(b) Rotor current flow from segment 2 to 1 (slot b to a)
20
Generator
21
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
22
DC Generator Equivalent circuit
Rf
Vbrush
max
Vf If
Eag
Mechanical
power in
Ra
Iag
Load
Vdc
Electrical
power out
• Figure 8.12Equivalent circuit of a separately excited dc
generator.
23
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:
 ag  K 1 I f
24
DC Generator Equivalent circuit
• The rotor conductors cut the field lines that
generate voltage in the coils.
Eag  2 N r B  g v
• The motor speed and flux equations are :
v 
Dg
2
 ag  B  g D g
25
DC Generator Equivalent circuit
• The combination of the three equation
results the induced voltage equation:
Eag
 Dg 
  N r B  g Dg   N r  ag 
 2 N r B  g v  2 N r B  g 
2 

• The equation is simplified.
E ag  N r  ag   N r K1 I f   K m I f 
26
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;
E ag  Vdc  I ag Ra  Vbrush
27
Motor
28
DC Motor Equivalent circuit
Rf
Vbrush
Ra
Electrical
power in
max
Vf If
Eam
Iam
Vdc
DC Power
supply
Mechanical
power out
• Figure 8.13 Equivalent circuit of a separately excited dc motor
• Equivalent circuit is similar to the generator only the current
directions are different
29
DC Motor Equivalent circuit
• The operation equations are:
• Armature voltage equation
Vdc  Eam  I am Ra  Vbrush
The induced voltage and motor speed vs angular
frequency
E am  K m I f 
  2  nm
30
DC Motor Equivalent circuit
• The operation equations are:
• The combination of the equations results in
K m I f   E am  Vdc  I am Rm
The current is calculated from this equation. The output
power and torque are:
Pout  Eam I am
T
Pout

 K m I am I f
31
Ideal Transformers
32
The ideal transformer:
I1
+
- V1
IT
I2
V2
N1
I2
I1
+
-
V1
N2
V2
V1/N1 = V2/N2
N 1 I1 = N 2 I2
S 1 = V 1 I1 *
S 2 = V 2 I2 *
S 1 = S2
k = V2/V1 = N2/N1
V1 = V2/k
I1
I2
V1
V2
N1
Z2 = V2/I2
I1 = k* I2
Z2
N2
Z2’ = V1/I1 = (V2/k) / (I2*k) = Z2/k2
Real transformer:
I1
R1 jX1
Ic
V1
R0
R2
I2*k
Im
V1
V2
jX2 I2
Load
jX0
N1
N2
33
+
V1
V2
V2
N1 IT N2
k = V2
V1
I1
=
N2
N1
R1
jX1
Ic
V1
Rc
I2*k
I2
R2
Im
Load
jXm
N1
I1
V1
R1
jX2
N2
R2’ jX2’
jX1
Ic
Im
Rc
jXm
I2
N1
N2
34
I1
+
- V1
IT
R1jX1
I1
V2
V2
N2

V1
N1
Im
Rc
I2
V1
N2
R2
I2*k
Ic
V1
I1
+
-
V2
N1
k 
I2
V1
jX2 I2
Load
V2
jXm
N1
N2
R2 & X2 are referred to
winding 1
I1
R1
Ic
V1
R2’ jX2’
jX1
Rc
I2
Im
Load
jXm
N1
N2
35
I1
Req
jXeq
I2
V2
V1
Load
N1
N2
Req = R1 + R2’
Xeq = X1 + X2’
I1
Neglecting excitation
current
jXeq
I2
V1
V2
Load
N1
N2
36
Determination of Req and Xeq from short-circuit test.
Isc
Vsc
Req
jXeq
Vsc
Ideal Transformer
Real Transformer
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
Isc
Vsc
Isc Req jXeq
Vsc
Vsc = .05 x V2rated
Measure Isc, Psc, Vsc
The same procedure as before however, the short-circuit impedance is referred to 37
winding #2.
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
38
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 
39
I1
Rs
jXs
I2
V1(rated)
LV
V2(rated)
N1
N2
S(rated)
Rs’
I1
V1(rated)
HV
N1
a=
N2
=
V1(rated)
jXs’
V2(rated)
N1
N2
S(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 = Vb12/Sb Sb = Vb1Ib1
Zb2 = Vb2/Ib2 = Vb22/Sb
Ib2 = Sb/ Vb2
Zb1 / Zb2 = Vb12/ Vb22 = Vb12/( 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
40
Rs
jXs
V1(rated)
V2(rated)
N1
N2
Vb1
Vb1 = V1(rated) Sb = S(rated 1 phase)
Vb2 = V2(rated)
Zb1 = (Vb1)2/Sb Zb2 = (Vb2)2/Sb
I1 pu
V1 pu
Zpu
Zpu
Vb1
Vb2
Vb1
Zpu = Rs+jXs/Zb1
Rs/a2 + jXs/a2
V2(rated)
N1
Vb2
N2
Vb2
Sb
Vb1
I1 pu
V1 pu
Vb2
I2 pu
V2 pu
Sb = S1(rated) Vb2 = V2(rated)
Zb1 = (Vb1)2/Sb Zb2 = (Vb2)2/Sb
Zb2 = Zb1*k2
41
Transformation between two bases:
ZL
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))
42