DC Machines with explanation in detail of everything

DC Machines with explanation in detail of everything

1. Chapter 6
DC Machines
EET103/4

2. Introduction
•An electrical machine is link between an electrical
system and a mechanical system.
•Conversion from mechanical to electrical: generator
•Conversion from electrical to mechanical: motor

3. Introduction
Machines are called
•AC machines (generators or motors) if the electrical
system is AC.
•DC machines (generators or motors) if the
electrical system is DC.

4. DC machines can be divide by:
a) DC motor
b) DC Generator
DC Machines
DC Motor DC Generator

5. DC Machines Construction
cutaway view of a dc machine

6. DC Machines Construction
cutaway view of a dc machine

7. DC Machines Construction
Rotor of a dc machine

8. DC Machines Construction
Stator of a dc machine

9. DC Machines Fundamentals
•Stator: is the stationary part of the machine. The
stator carries a field winding that is used to
produce the required magnetic field by DC
excitation.
•Rotor (Armature): is the rotating part of the
machine. The rotor carries a distributed winding,
and is the winding where the e.m.f. is induced.
•Field winding: Is wound on the stator poles to
produce magnetic field (flux) in the air gap.
•Armature winding: Is composed of coils placed in
the armature slots.
•Commutator: Is composed of copper bars,
insulated from each other. The armature winding is
connected to the commutator.
•Brush: Is placed against the commutator surface.
Brush is used to connect the armature winding to
external circuit through commutator

10. DC Machines Fundamentals
In DC machines, conversion of energy from electrical to mechanical
form or vice versa results from the following two electromagnetic
phenomena
1.When a conductor moves in a magnetic field, voltage is induced in
the conductor.
2. When a current carrying conductor is placed in magnetic field, the
conductor experiences a mechanical forces.

11. DC Machines Fundamentals
Generator action:
An e.m.f. (voltage) is induced in a conductor if it
moves through a magnetic field.
Motor action:
A force is induced in a conductor that has a current
going through it and placed in a magnetic field
•Any DC machine can act either as a generator or
as a motor.

12. DC Machines Equivalent Circuit
The equivalent circuit of DC machines has two
components:
Armature circuit:
•It can be represented by a voltage source and a
resistance connected in series (the armature
resistance). The armature winding has a
resistance, R
A.
The field circuit:
•It is represented by a winding that generates the
magnetic field and a resistance connected in
series. The field winding has resistance R
F.

13. DC Motor

14. •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.
Basic Operation of DC Motor

15. •The magnetic field lines enter into
the rotor from the north pole (N)
and exit toward the south pole (S)
•The poles generate a magnetic
field that is perpendicular to the
current carrying conductors
•The interaction between the field
and the current produces a
Lorentz force
•The force is perpendicular to both
the magnetic field and conductor
Basic Operation of DC Motor

16. Basic Operation of DC Motor
•The generated force turns the
rotor until the coil reaches the
neutral point between the poles.
•At this point, the magnetic field
becomes practically zero
together with the force.
•However, inertia drives the motor
beyond the neutral zone where
the direction of the magnetic field
reverses.
•To avoid the reversal of the force
direction, the commutator
changes the current direction,
which maintains the counter
clockwise rotation.

17. •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
Basic Operation of DC Motor

18. Basic Operation of DC Motor

19. Classification of DC Motor
1. Separately Excited DC Motor
•Field and armature windings are either connected
separate.
2. Shunt DC Motor
•Field and armature windings are either connected in
parallel.
3. Series DC Motor
•Field and armature windings are connected in
series.
4. Compound DC Motor
•Has both shunt and series field so it combines
features of series and shunt motors.

20. Important terms
•V
T–supply voltage
•E
A–internal generated voltage/back e.m.f.
•R
A–armature resistance
•R
F–field/shunt resistance
•R
S–series resistance
•I
L–load current
•I
F–field current
•I
A–armature current
•I
L–load current
•n–speed

21. Generated or back e.m.f. of DC
Motor
•General form of back e.m.f.,
Φ= flux/pole (Weber)
Z= total number of armature conductors
= number of slots x number of conductor/slot
P= number of poles
A= number of parallel paths in armature
[A= 2 (for wave winding), A = P (for lap winding)]
N= armature rotation (rpm)
E
A= back e.m.f. A
PZN
E
A 
60


22. Torque Equation of a DC Motor
•The armature torque of a DC motor is given by
Φ= flux/pole (Weber)
Z = total number of armature conductors
= number of slots x number of conductor/slot
P= number of poles
A= number of parallel paths in armature
I
A= armature current
T
a= armature torque)(
2
meterNewton
A
PZI
T
A
a 



23. Equivalent Circuit of DC MotorF
T
F
R
V
I AAAT RIEV  F
F
F
R
V
I AAAT RIEV  ALII
Separately Excited DC Motor
Shunt DC MotorFAL III 

24. )(
SAAAT RRIEV  LSA III  Series DC Motor)(
SAAAT RRIEV  F
T
F
R
V
I FLA III 
Compound DC Motor

25. Speed of a DC Motor
•For shunt motor
•For series motor1
2
1
2
12
2
1
1
2
1
2
,
A
A
A
A
E
E
n
n
thenIf
E
E
n
n
 

 2
1
1
2
2
1
1
2
1
2
A
A
A
A
A
A
I
I
E
E
E
E
n
n




26. Example 1
A 250 V, DC shunt motor takes a line
current of 20 A. Resistance of shunt
field winding is 200 Ωand resistance of
the armature is 0.3 Ω. Find the
armature current, I
Aand the back e.m.f.,
E
A.

27. Solution
Given quantities:
•Terminal voltage, V
T = 250 V
•Field resistance, R
F = 200 Ω
•Armature resistance, R
A = 0.3 Ω
•Line current, I
L= 20 A
Figure 1

28. Solution (cont..)
the field current,
the armature current,
V
T= E
A+ I
AR
A
the back e.m.f.,
E
A= V
T–I
AR
A= 250 V –(18.75)(0.3) = 244.375 VA25.1
200
V250




F
T
F
FAL
R
V
I
III 18.75A
A25.1A20



FLA III

29. Example 2
A 50hp, 250 V, 1200 r/min dc shunt motor
with compensating windings has an
armature resistance (including the
brushes, compensating windings, and
interpoles) of 0.06 Ω. Its field circuit has a
total resistance R
adj + R
Fof 50 Ω, which
produces a no-load speed of 1200 r/min.
There are 1200 turns per pole on the
shunt field winding.

30. Example 2 (cont..)
a)Find the speed of this motor when its
input current is 100 A.
b)Find the speed of this motor when its
input current is 200 A.
c)Find the speed of this motor when its
input current is 300 A.

31. Solution
Given quantities:
•Terminal voltage, V
T = 250 V
•Field resistance, R
F = 50 Ω
•Armature resistance, R
A = 0.06 Ω
•Initial speed, n
1= 1200 r/min
Figure 2

32. Solution (cont..)
(a) When the input current is 100A, the armature
current in the motor is
Therefore, E
Aat the load will beA95A5A100
50
V250
A100




F
T
LFLA
R
V
IIII V3.244
V7.5V250
)06.0)(A95(V250




AATA RIVE

33. Solution(cont..)
•The resulting speed of this motor ismin/r1173
min/r1200
250
3.244
1
1
2
2
1
2
1
2




V
V
n
E
E
n
E
E
n
n
A
A
A
A

34. Solution(cont..)
(b) When the input current is 200A, the armature
current in the motor is
Therefore, E
Aat the load will beA195A5A200
50
V250
A200




F
T
LFLA
R
V
IIII V3.238
V7.11V250
)06.0)(195(V250




A
RIVE
AATA

35. Solution(cont..)
•The resulting speed of this motor ismin/r1144
min/r1200
250
3.238
1
1
2
2
1
2
1
2




V
V
n
E
E
n
E
E
n
n
A
A
A
A

36. Solution (cont..)
(c) When the input current is 300A, the armature
current in the motor is
Therefore, E
Aat the load will beA295A5A300
50
V250
A300




F
T
LFLA
R
V
IIII V3.232
V7.17V250
)06.0)(295(V250




A
RIVE
AATA

37. Solution (cont..)
•The resulting speed of this motor ismin/r1115
min/r1200
V250
V3.232
1
1
2
2
1
2
1
2




n
E
E
n
E
E
n
n
A
A
A
A

38. Example 3
The motor in Example 2 is now connected in
separately excited circuit as shown in Figure 3.
The motor is initially running at speed, n= 1103
r/min with V
A= 250 V and I
A= 120 A, while
supplying a constant-torque load. If V
Ais reduced
to 200 V, determine
i). the internal generated voltage, E
A
ii). the final speed of this motor, n
2

39. Example 3 (cont..)
Figure 3

40. Solution
Given quantities
•Initial line current, I
L= I
A= 120 A
•Initial armature voltage, V
A= 250 V
•Armature resistance, R
A= 0.06 Ω
•Initial speed, n
1= 1103 r/min

41. Solution (cont..)
i) The internal generated voltage
E
A=V
T-I
AR
A
= 250 V –(120 A)(0.06 Ω)
= 250 V –7.2 V
= 242.8 V

42. Solution (cont..)
ii) Use KVL to find E
A2
E
A2=V
T-I
A2R
A
Since the torque is constant ant he flux is constant,
I
Ais constant. This yields a voltage of
E
A2= 200 V –(120 A)(0.06 Ω)
= 200 V –7.2 V
= 192.8 V

43. Solution (cont..)
•The final speed of this motormin/r876
min/r1103
V8.242
V8.192
1
1
2
2
1
2
1
2




n
E
E
n
E
E
n
n
A
A
A
A

44. Example 4
A DC series motor is running with a speed
of 800 r/min while taking a current of 20 A
from the supply. If the load is changed
such that the current drawn by the motor
is increased to 50 A, calculate the speed
of the motor on new load. The armature
and series field winding resistances are
0.2 Ωand 0.3 Ωrespectively. Assume the
flux produced is proportional to the
current. Assume supply voltage as 250 V.

45. Solution
Given quantities
•Supply voltage, V
T = 250 V
•Armature resistance, R
A = 0.2 Ω
•Series resistance, R
S = 0.3 Ω
•Initial speed, n
1= 800 r/min
•Initial armature current, I
a1 =I
L1= 20 A
Figure 4

46. Solution (cont..)
For initial load, the armature current, I
a1 = 20 A and
the speed n
1= 800 r/min
V= E
A1+ I
a1(R
A + R
S)
The back e.m.f. at initial speed
E
A1= V -I
a1(R
A + R
S)
= 250 –20(0.2 + 0.3)
= 240 V

47. Solution (cont..)
When the armature current increased, I
a2 = 50 A, the
back emf
E
A2= V –I
a2(R
A + R
S)
= 250 –50(0.2 + 0.3)
= 225 Vmin/r300
50
20
240
225
800
2
1
1
2
12
2
1
1
2
1
2
2
1
1
2
1
2





I
I
E
E
nn
I
I
E
E
n
n
E
E
n
n
A
A
A
A
A
A


The speed of the motor on new load

48. DC Generator

49. Generating of an AC Voltage
•The voltage generated
in any dc generator
inherently alternating
and only becomes dc
after it has been rectified
by the commutator

50. Generation of an AC Voltage

51. Armature windings
•The armature windings are usually former-
wound. This are first wound in the form of flat
rectangular coils and are then puller.
•Various conductors of the coils are insulated
each other. The conductors are placed in the
armature slots which are lined with tough
insulating material.
•This slot insulation is folded over above the
armature conductors placed in the slot and is
secured in place by special hard wooden or fiber
wedges.

52. Lap and wave Windings
There are two types of windings mostly
employed:
• Lap winding
• Wave winding
The difference between the two is merely
due to the different arrangement of the end
connection at the front or commutator end of
armature.

53. Generated or back e.m.f. of DC
Generator
•General form of generated e.m.f.,
Φ = flux/pole (Weber)
Z = total number of armature conductors
= number of slots x number of conductor/slot
P = number of poles
A = number of parallel paths in armature
[A = 2 (for wave winding), A = P (for lap winding)]
N = armature rotation (rpm)
E = e.m.f. induced in any parallel path in armatureA
PZN
E 
60


54. Classification of DC Generator
1. Separately Excited DC Generator
•Field and armature windings are either connected
separate.
2. Shunt DC Generator
•Field and armature windings are either connected in
parallel.
3. Series DC Generator
•Field and armature windings are connected in
series.
4. Compound DC Generator
•Has both shunt and series field so it combines
features of series and shunt motors.

55. Equivalent circuit of DC generatorFAL III  ALII
Separately excited DC generatorF
F
F
R
V
I AAAT RIEV 
Shunt DC generatorF
T
F
R
V
I AAAT RIEV 

56. FAL III  ASL III  Series DC generator)(
SAAAT RRIEV 
Compound DC generatorF
T
F
R
V
I AAAT RIEV 

57. Example
•A DC shunt generator
has shunt field winding
resistance of 100Ω. It
is supplying a load of
5kW at a voltage of
250V. If its armature
resistance is 0.02Ω,
calculate the induced
e.m.f. of the generator.

58. Solution
Given quantities
•Terminal voltage, V
T = 250V
•Field resistance, R
F = 100Ω
•Armature resistance, R
A = 0.22Ω
•Power at the load, P= 5kW

59. Solution (cont..)A5.2
100
V250




F
T
F
FLA
R
V
I
III
The field current,A20
V250
W5000

T
L
V
P
I
The load current,
The armature current, I
A= I
L+ I
F= 20A + 2.5A = 22.5A
The induced e.m.f.,
E
A= V
T+ I
A R
A= 250V + (22.5)(0.22) = 254.95V