1. Electromechanical Systems
Unit 7
DC Motors 1

2. Summary and Learning Outcomes
After completing this unit you:
Will learn about the operation of direct current motors, different types of DC motors and their applications
Should be able to calculate
Analyse the performance and principle of operation of DC motors.
Calculate the torque speed characteristics for: -
Shunt wound DC motors
Series wound DC motors
Separately excited DC motors

3. Content DC Motor Construction Methods of Connection
Principle of Operation
Induced emf in the Armature
Torque
Armature Terminal Voltage
Methods of Connection
Shunt Wound DC Motor
Series Wound DC Motor
Separately Excited DC Motor

4. The World’s 1st Machine

5. The World’s 1st Machine
Faraday’s Magic

6. Some Machine Terminology
Electric machines can be classified in terms of their energy conversion characteristics.
Generators convert mechanical energy from a prime mover (e.g., an internal combustion engine) to electrical form.
Examples of generators are those used in power-generating plants, or automotive alternator.
Motors convert electrical energy to mechanical form.
Electric motors provide forces and torques to generate motion in countless industrial applications.
For Example Machine tools, robots, punches, presses, mills, and propulsion systems for electric vehicles are but a few examples of the application of electric machines in engineering.

7. Some Windings Terminology
Distinction can be made between different types of windings characterized by the nature of the current they carry.
If the current serves the purpose of providing a magnetic field and is independent of the load, (it is called a magnetizing, or excitation, current) the winding is termed a field winding.
(nearly always DC and are of relatively low power, since their only purpose is to magnetize the core).
However, if the winding carries only the load current, it is called an armature.
In DC and AC synchronous machines, separate windings exist to carry field and armature currents.

8. Motors/Generators Construction
A Motor/Generator are made of
Stator: This is the stationary part
Rotor: This is the rotating part
Separated by an air gap

9. Motors/Genrators Construction
The rotor and stator each consist of
Magnetic core,
Electrical insulation, and
Windings necessary to establish a magnetic flux (unless this is created by a permanent magnet).
The rotor is mounted on a bearing-supported shaft, which can be connected to:
Mechanical loads (functioning as a motor), or
A prime mover (functioning as a generator) by means of belts, pulleys, chains, or other mechanical couplings.
The windings carry electric currents that generate magnetic fields
(by virtue of Faraday’s law)

10. + = Partially wounded Motors Construction Commutator
DC Machine
Rotor:
Armature conductor are connected to the Commutator
Commutator
Mechanical rectifier converts ac to dc
Made of copper segment insulated by mica
Brushes
Electrical connector between armature and power
Pressure is adjusted using the spring
Stator
Produces an external flux

11. Principle of Operation
When a current carrying conductor is placed in a magnetic field, the conductor experience a mechanical force.
Direction is given by Flemings left hand rule
( F- B; S-I; T- M)
Magnitude is F=B.I.L
Consider a motor with one pair of poles, an armature with a single conductor coil and a commutator with only two segments,
If is field current supplied to the field winding to establish the main field between the poles N and S.
Ia is armature current via the carbon brushes. This current produces magnetic fields around the armature conductors
Explanation

12. Fleming Left Hand Rule Magnitude is F=BIL
Magnetic field due to Stator and Filed
Stator and Filed Magnetic interaction

13. Fleming Left Hand Rule

14. Fleming Left Hand Rule

15. Fleming Left Hand Rule

16. Fleming Left Hand Rule

17. Fleming Left Hand Rule

18. Fleming Left Hand Rule

19. Fleming Left Hand Rule

20. Fleming Left Hand Rule

21. Fleming Left Hand Rule

22. Fleming Left Hand Rule

23. Fleming Left Hand Rule

24. Fleming Left Hand Rule

25. Fleming Left Hand Rule

26. Induced EMF in the Armature
As the coil rotates an emf is induced in each conductor which opposes the externally supplied armature current, Ia.
The external supply must overcome this emf if the machine is to continue motoring and deliver mechanical power through its shaft.
Faraday’s Law states that the
emf induced in a conductor = rate of change of flux linkages
Taken over a period of time
Average emf induced in conductor = total flux linkage
total time of linkage
So, When conductor 1 is close to N-pole:
Total flux emanating from that pole = 
Average emf induced in conductor1 = total flux linkage
total time of linkage
Therefore, if the coil rotating at n rev sec-1
Each conductor will be close to a particular pole 2n times per rotation
Each conductor will link with its magnetic flux for sec per rotation 26

27. E = 2pn  As E = kn Volts Number of poles affects the induced emf
Machines have several pairs of poles.
For a machine with p pole pairs, the average emf in each conductor is given by:
average emf induced = total flux per pole
in a conductor total time conductor is under a pole
Total emf induced = average emf induced  number of armatur
in armature winding in one conductor conductors in series
E = 2pn  As
The number of poles (2p) and the number of armature conductors in series (As) are constant for a particular machine. Therefore k = 2p As
E = kn Volts
Since the angular velocity,  = 2n

28. The “lost” torque is small and will be ignored
Electrical power delivered = Armature emf  Armature current
to the armature
Pa = E  Ia
This power creates the torque to make the armature rotate.
Electrical torque developed = Electrical power delivered to the armature
in the armature Angular velocity
Remember: Power is the rate at which work is done; Work done in 1 s = force × distance
Power = work done / time taken
Mechanical torque|at the shaft = Electrical torque - “Lost” torque|due to frictional and other losses
The “lost” torque is small and will be ignored

29. Armature Terminal Voltage
The figure represents an equivalent circuit of an armature
E is the induced emf
Ra is the armature resistance
The armature terminal voltage is given by:
Va = E + IaRa

30. Methods of Connection
The field and armature windings may be connected to:
Independent supplies - separately excited
Common supply - self excited
Shunt wound: The field and armature windings are connected in parallel
Series wound: The field and armature windings are in series
Compound wound: Has two field windings;
One connected parallel with the armature and
Other in series with the armature

31. hence keeping If and hence  constant.
Shunt Wound DC Motors
Field Winding:
Where, S is the reluctance,
N is the number of turns in the coil and
i is the coil current.
Armature Winding:
Armature terminal voltage, V = E + IaRa
V = kn + IaRa
with  constant, let K1 = k V = K1n + IaRa
V : External supply voltage
Rf : field winding resistance
are normally constant,
hence keeping If and hence  constant.

32. Speed Current Characteristics
From Last Slide

33. Torque Current Characteristics

34. Therefore, ideal for use with machine tools, pumps, compressors etc.
Torque Speed Characteristics
We had:
The torque-speed curve shows that shunt motors can be used to drive fairly constant speed from no load to full torque
Therefore, ideal for use with machine tools, pumps, compressors etc.

35. Exercise
A 220V dc shunt motor has an armature resistance of 0.8 and field winding resistance of 220. The motor field characteristic [k versus field current] is shown in Figure
a) Calculate the field current
If the motor drives a constant load torque of 17.5Nm, calculate
b) armature current
c) speed

36. Solution
Speed = 37.1 rev sec-1
= 2225 rev min-1.

37. Series Wound DC Motors
In the series motor current, I flows through both field and armature windings so:
V = E + I(Ra + Rf)
let R = Ra + Rf  V = E + IR
 E = V - IR

38. Speed Current Characteristics
All dc motors, flux,   field winding current
For Series wound motor   I;  = K3I
N.B. this assumption only applies for low currents.
E = kn
E = kK3In
E = K4In where K4 = kK3
 V = K4.I.n + I.R

39. Torque Current and Torque Speed Characteristics
The series motor is a variable speed machine ideally suited to drive permanently coupled loads.
They are often used for electric traction and lifts. They must never be used on “no load” as the speed will become dangerously high.

40. Exercise
A 220V dc series motor has armature and field resistances of 0.2 and 0.5 respectively. When running at 1000 rev min-1 the motor draws 10A from the supply. Calculate the torque delivered.
Solution

41. Separately Excited dc Motors
For accurate speed control it is advisable to use a separately excited motor
i.e. Armature and Field Windings supplied through independent dc rectifiers
The diode rectifier supplies constant field current maintaining a fixed value of flux, .
The controlled rectifier (supplying the armature winding) provides a fully variable armature terminal voltage, Va.

42. Separately Excited dc Motors, Equivalent Circuit
Va = E + IaRa But, E = k..n  Va = kn + IaRa
Ra is usually small so Va > IaRa. Thus with  constant the speed n, is almost directly proportional to Va.
Used for accurate speed control.

43. Conclusion Today we learnt about DC motors
The three types of DC motors
Shunt wound DC motors
Series wound DC motors
Separately excited DC motor
and their applications
We also touched on how to:
Analyse the performance and principle of operation of DC motors.
Calculate the torque speed characteristics for the three different types

44. Problems
Q1 A 240V dc shunt motor has armature and field resistances 0.2  and 320  respectively. The motor drives a load at a speed of 950 rev min-1 and the armature current is 50A. Assuming that the flux is directly proportional to the field current, calculate the additional resistance necessary in the field circuit to increase the speed to 1100 rev min-1 while maintaining the armature current constant.
Calculate the speed of the machine with the original field current and an armature current of 90A.
50.5 , 917 rev min-1
Q2 A 230V dc shunt motor has armature and field resistances of 0.3  and 140  respectively. Calculate the induced emf and the torque developed by the motor when it runs at a speed of 800 rev min-1 and the armature current is 2A.
To drive a larger load at 1000 rev min-1 an additional resistance, R is connected in series with the field winding. In this situation the armature current is 30A. Calculate the new induced emf and torque and the value of R. Assume that the flux is directly proportional to the field current.
229.4V, 5.48Nm; 221V, 63.3Nm, 41.7 
Q3 A 240V dc series motor has armature and field resistances of 0.5  and 1  respectively. When running at 1200 rev min-1 the motor draws 15A from the supply. Calculate the torque delivered.
A 2  resistor is connected in series with the motor. The torque is adjusted so that the armature current remains unchanged. Calculate the new speed and torque. 26Nm; 1034 rev min-1; 26Nm
Q4 A 550V dc series motor with an armature resistance of 0.35  and and field resistance of 0.15  drives a load at a speed of 750 rev min-1. The supply current is 74A. Calculate the load torque.
The load torque is doubled and the supply current rises to 110A. Calculate the new speed and power output Nm; rev min-1; kW