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

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
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, RA.
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 RF.

13. DC Motor

14. Basic Operation of DC Motor
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.

15. Basic Operation of DC Motor
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

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. Basic Operation of DC Motor
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

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 VT – supply voltage
EA – internal generated voltage/back e.m.f.
RA – armature resistance
RF – field/shunt resistance
RS – series resistance
IL – load current
IF – field current
IA – armature 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)
EA = back e.m.f.

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
IA = armature current
Ta = armature torque

23. Equivalent Circuit of DC Motor
Separately Excited DC Motor
Shunt DC Motor

24. Series DC Motor
Compound DC Motor

25. Speed of a DC Motor
For shunt motor
For series motor

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

27. Solution Given quantities: Terminal voltage, VT = 250 V
Field resistance, RF = 200 Ω
Armature resistance, RA = 0.3 Ω
Line current, IL = 20 A
Figure 1

28. Solution (cont..) the field current, the armature current,
VT = EA + IARA
the back e.m.f.,
EA = VT – IARA = 250 V – (18.75)(0.3) = V

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 Radj + RF of 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..)
Find the speed of this motor when its input current is 100 A.
Find the speed of this motor when its input current is 200 A.
Find the speed of this motor when its input current is 300 A.

31. Solution Given quantities: Terminal voltage, VT = 250 V
Field resistance, RF = 50 Ω
Armature resistance, RA = 0.06 Ω
Initial speed, n1 = 1200 r/min
Figure 2

32. Solution (cont..)
(a) When the input current is 100A, the armature current in the motor is
Therefore, EA at the load will be

33. Solution (cont..)
The resulting speed of this motor is

34. Solution (cont..)
(b) When the input current is 200A, the armature current in the motor is
Therefore, EA at the load will be

35. Solution (cont..)
The resulting speed of this motor is

36. Solution (cont..)
(c) When the input current is 300A, the armature current in the motor is
Therefore, EA at the load will be

37. Solution (cont..)
The resulting speed of this motor is

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 VA = 250 V and IA = 120 A, while supplying a constant-torque load. If VA is reduced to 200 V, determine
i). the internal generated voltage, EA
ii). the final speed of this motor, n2

39. Example 3 (cont..)
Figure 3

40. Solution Given quantities Initial line current, IL = IA = 120 A
Initial armature voltage, VA = 250 V
Armature resistance, RA = 0.06 Ω
Initial speed, n1 = 1103 r/min

41. Solution (cont..) i) The internal generated voltage EA = VT - IARA
= 250 V – (120 A)(0.06 Ω)
= 250 V – 7.2 V
= V

42. Solution (cont..) ii) Use KVL to find EA2 EA2 = VT - IA2RA
Since the torque is constant ant he flux is constant, IA is constant. This yields a voltage of
EA2 = 200 V – (120 A)(0.06 Ω)
= 200 V – 7.2 V
= V

43. Solution (cont..)
The final speed of this motor

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 Ω and 0.3 Ω respectively. Assume the flux produced is proportional to the current. Assume supply voltage as 250 V.

45. Solution Given quantities Supply voltage, VT = 250 V
Armature resistance, RA = 0.2 Ω
Series resistance, RS = 0.3 Ω
Initial speed, n1 = 800 r/min
Initial armature current, Ia1 = IL1 = 20 A
Figure 4

46. Solution (cont..)
For initial load, the armature current, Ia1 = 20 A and the speed n1 = 800 r/min
V = EA1 + Ia1 (RA + RS)
The back e.m.f. at initial speed
EA1 = V - Ia1 (RA + RS)
= 250 – 20( )
= 240 V

47. The speed of the motor on new load
Solution (cont..)
When the armature current increased, Ia2 = 50 A, the back emf
EA2 = V – Ia2 (RA + RS)
= 250 – 50( )
= 225 V
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 armature

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 generator
Separately excited DC generator
Shunt DC generator

56. Series DC generator
Compound DC generator

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, VT = 250V
Field resistance, RF = 100Ω
Armature resistance, RA = 0.22Ω
Power at the load, P = 5kW

59. Solution (cont..) EA = VT + IA RA = 250V + (22.5)(0.22) = 254.95V
The field current,
The load current,
The armature current, IA = IL + IF = 20A + 2.5A = 22.5A
The induced e.m.f.,
EA = VT + IA RA = 250V + (22.5)(0.22) = V