1. Professor Mohammed Zeki Khedher
Electric Drives
Professor Mohammed Zeki Khedher
Lecture One

2. Introduction
In some countries nearly 65% of the total electric energy produced is consumed by electric motors.

3. Some Applications of Electric Drives
Electric Propulsion
Pumps, fans, compressors
Plant automation
Flexible manufacturing systems
Spindles and servos
Appliances and power tools
Cement kilns
Paper and pulp mills; textile mills
Automotive applications
Conveyors, elevators, escalators, lifts

4. 1. ENERGY CONVERSION IN ELECTRIC DRIVES
1.1. ELECTRIC DRIVES - A DEFINITION
About 50% of electrical energy produced is used in electric drives today.
Electric drives may run at constant speed (figure 1.1) or at variable speed (figure 1.2).
Figure 1.1. Constant speed electric drive

5. Figure 1.2. Variable speed electric drive

6. 1.2. APPLICATION RANGE OF ELECTRIC DRIVES
A summary of main industrial applications and power range of electric drives is shown on figure 1.3.
Figure 1.3. Electric drives - variable speed applications

7. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives 7

8. Energy/Cost Savings
System efficiency can be increased from 15% to 27% by introducing variable-speed drive operation in place of constant-speed operation.
For a large pump variable-speed drive, payback period ~ 3-5 years whereas operating life is ~ 20 years.

9. Electric Machines
“An engineer designing a high-performance drive system must have intimate knowledge about machine performance and Power Electronics”

10. Electric Machines (cont’d)
DC Machines - shunt, series, compound, separately excited dc motors and switched reluctance machines
AC Machines - Induction, wound rotor synchronous, permanent magnet synchronous, synchronous reluctance, and switched reluctance machines.
Special Machines - switched reluctance machines

11. Electric Machines (cont’d)
All of the above machines are commercially available in fractional kW to MW ranges except permanent-magnet, synchronous, synchronous reluctance, and switched reluctance which are available up to 150 kW level.

12. Selection Criteria for Electric Machines
Cost
Thermal Capacity
Efficiency
Torque-speed profile
Acceleration
Power density, volume of motor
Ripple, cogging torques
Peak torque capability

13. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Electrical Drives
About 50% of electrical energy used for drives
Can be either used for fixed speed or variable speed
75% - constant speed, 25% variable speed (expanding) 13

14. Example on VSD application
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
Constant speed
Variable Speed Drives
motor
pump
valve
Supply
Power In
Power loss
Mainly in valve
Power out

15. Example on VSD application
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
Constant speed
Variable Speed Drives
motor
pump
valve
Supply
motor
PEC
pump
Supply
Power loss
Mainly in valve
Power out
Power In
Power In
Power loss
Power out

16. Example on VSD application
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
Constant speed
Variable Speed Drives
motor
pump
valve
Supply
motor
PEC
pump
Supply
Power In
Power loss
Mainly in valve
Power out
Power loss
Power In
Power out

17. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Conventional electric drives (variable speed)
Bulky
Inefficient
inflexible 17

18. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Modern electric drives (With power electronic converters)
Small
Efficient
Flexible 18

19. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Modern electric drives
Non-linear control
Real-time control
DSP application
PFC
Speed sensorless
Power electronic converters
Utility interface
Renewable energy
Machine design
Speed sensorless
Machine Theory
Inter-disciplinary
Several research area
Expanding 19

20. Controllers
Controllers embody the control laws governing the load and motor characteristics and their interaction.
Controller
Torque/speed/
position commands
Vc, fc, start,
shut-out,
signals, etc.
Torque/speed/
position feedback
Thermal and
other feedback

22. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
e.g. Single drive - sensorless vector control from Hitachi 22

23. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
e.g. Multidrives system from ABB 23

24. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
Motors
DC motors - permanent magnet – wound field
AC motors – induction, synchronous , brushless DC
Applications, cost, environment
Power sources
DC – batteries, fuel cell, photovoltaic - unregulated
AC – Single- three- phase utility, wind generator - unregulated
Power processor
To provide a regulated power supply
Combination of power electronic converters
More efficient
Flexible
Compact
AC-DC DC-DC DC-AC AC-AC 24

25. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
Control unit
Complexity depends on performance requirement
analog- noisy, inflexible, ideally has infinite bandwidth.
digital – immune to noise, configurable, bandwidth is smaller than the analog controller’s
DSP/microprocessor – flexible, lower bandwidth - DSPs perform faster operation than microprocessors (multiplication in single cycle), can perform complex estimations 25

26. DC Motors
Advantage: simple torque and speed control without sophisticated electronics
Limitations:
Regular Maintenance
Expensive motor
Heavy motor
Sparking

27. DC DRIVES Vs AC DRIVES DC drives: Advantage in control unit
Disadvantage in motor
AC Drives:
Advantage in motor
Disadvantage in control unit

28. 1.3. ENERGY SAVINGS PAYS OFF RAPIDLY
Consider a real case when a motor pump system of 15kW works 300 days a year, 24 hours a day and pumps 1200m3 of water per day. By on/off and throttling control, only, the system uses 0.36kWh/m3 of pumped water to keep the pressure rather constant for variable flow rate.
Adding a P.E.C., in the same conditions, the energy consumption is 0.28kWh/m3 of pumped water, with a refined pressure control.
Let us consider that the cost of electrical energy is 40fils/kWh.
The energy savings per year S is:
S = 1200 * 300 * ( ) * 0.04 /year = 1152 JD/year
Now the costs of a 15kW PWM - P.E.C. for an induction motor is less than 4000JD. Thus, to a first approximation, the loss savings only pay off the extra investment in less than 4 years.

29. Costs
Power Electronics Controller costs approximately 2 to 5 times AC motor
Cost decreases with bigger size

30. 1.4. GLOBAL ENERGY SAVINGS THROUGH P.E.C. DRIVES
So far the energy savings produced by the P.E.C. in variable speed drives have been calculated for the drive only - P.E.C. and motor.
Figure 1.5. Primary energy consumption for throttle / motor / pump system

31. Figure 1. 6. Primary energy consumption for P. E. C
Figure 1.6. Primary energy consumption for P.E.C. / motor / pump systems

32. Power consumption with flow

33. Load
The motor drives a load that has a characteristic torque vs. speed requirement.
In general, load torque is a function of speed and can be written as:
Tl  mx
x=1 for frictional systems (e.g. feed drives)
x=2 for fans and pumps

34. General Torque Equation
Translational (linear) motion:
F : Force (Nm)
M : Mass (Kg )
v : velocity (m/s)
Rotational motion:
T : Torque (Nm)
J : Moment of Inertia (Kgm2 )
w : angular velocity ( rad/s )

35. Torque Equation: Motor drives
Te : motor torque (Nm)
TL : Load torque (Nm)
Acceleration
Deceleration
Constant speed

36. …continue
Drive accelerates or decelerates depending on whether Te is greater or less than TL
During acceleration, motor must supply not only
the load torque but also dynamic torque, ( Jdw/dt ).
During deceleration, the dynamic torque, ( Jdw/dt ), has
a negative sign. Therefore, it assists the motor
torque, Te.

37. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics v x
Newton’s law Fm M Ff
Linear motion, constant M
First order differential equation for speed
Second order differential equation for displacement 37

38. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics 
Te , m Tl
Rotational motion
- Normally is the case for electrical drives J
First order differential equation for angular frequency (or velocity)
Second order differential equation for angle (or position)
With constant J, 38

39. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
For constant J,
Torque dynamic – present during speed transient
Angular acceleration (speed)
The larger the net torque, the faster the acceleration is. 39

40. Elementary principles of mechanics
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
Combination of rotational and translational motions r 
Te,  Tl Fl Fe v M
Te = r(Fe), Tl = r(Fl), v =r
r2M - Equivalent moment inertia of the
linearly moving mass 40

41. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics – effect of gearing
Motors designed for high speed are smaller in size and volume
Low speed applications use gear to utilize high speed motors
Motor Te
Load 1, Tl1
Load 2, Tl2 J1 J2 m
m1
m2 n1 n2 41

42. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics – effect of gearing
Motor Te
Load 1, Tl1
Load 2, Tl2 J1 J2 m
m1
m2 n1 n2
Motor Te
Jequ
Equivalent Load , Tlequ m
Tlequ = Tl1 + a2Tl2
a2 = n1/n2 42

43. 1.6. MOTION / TIME PROFILE MATCH
Figure 1.8. Motion / time profile a.) speed b.) position c.) required load torque

44. Figure 1.9. Required speed / time profile
Example 1.2. The direct drive torque / time curve
A direct drive has to provide a speed / time curve such as in figure 1.9. against a constant load torque of TL = 10Nm, for a motor load inertia J = 0.02 kgm2.
Figure 1.9. Required speed / time profile
Neglecting the mechanical losses let us calculate the motor torque (Te) / time requirements.
The motion equation for a direct drive is:

45. For the linear speed / time (acceleration - deceleration) zones the speed derivative is:
For the constant speed (cruising) zone
Consequently the torque requirements from the motor for the three zones are:

46. Figure 1.10. Motor torque / time requirements
The motor torque / time requirements are shown on figure 1.10.
Figure Motor torque / time requirements

47. Example 1.3. gear - box drive torque / time curve
Let us consider an electric drive for an elevator with the data shown in figure 1.11.
Figure Elevator electric drive with multiple mechanical transmissions and counterweight

48. The motor rated speed nn = 1550rpm
The motor rated speed nn = 1550rpm. The efficiency of gearing system is h = 0.8.
Let us calculate the total inertia (reduced to motor shaft), torque and power without and with counterweight.
First the motor angular speed wm is:
(1.12)
The gear ratios may be defined as speed ratios - Wt /wm for J4+J5 and Wd /wm for J6 (figure 1.11).
Consequently the inertia of all rotating parts Jr, reduced to the motor shaft, (figure 1.11), is:
(1.13)

49. For the cabin and the counterweight, the inertia, reduced to motor shaft (Je) is:
(1.14)
Thus the total inertia Jt is:
(1.15)
In absence of counterweight the la of energy conservation leads to:
(1.16)
Consequently the motor torque, Tem, yields:
(1.17)

50. The motor electromagnetic power Pem is:
(1.18)
On the other hand in presence of counterweight (1.16) becomes:
(1.19)
(1.20)
So the motor electromagnetic P’em is:
(1.21)

51. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Motor steady state torque-speed characteristic
Synchronous mch
Induction mch
Separately / shunt DC mch
Series DC
SPEED
TORQUE
By using power electronic converters, the motor characteristic can be change at will 51

52. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Frictional torque (passive load)
Exist in all motor-load drive system simultaneously
In most cases, only one or two are dominating
Exists when there is motion
SPEED
T~ C
Coulomb friction
T~ 2
Friction due to turbulent flow
T~ 
Viscous friction
TORQUE 52

53. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Constant torque, e.g. gravitational torque (active load)
SPEED
TORQUE
Gravitational torque  TL Te
Vehicle drive FL gM
TL = rFL = r g M sin  53

54. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Hoist drive
Speed
Torque
Gravitational torque 54

55. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load and motor steady state torque
At constant speed, Te= Tl
Steady state speed is at point of intersection between Te and Tl of the steady state torque characteristics Tl Te
Torque
r2
r3
r1
Steady state
speed r
Speed 55

56. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile 10 25 45 60
t (ms)
speed
(rad/s)
100
Speed profile
The system is described by: Te – Tload = J(d/dt) + B
J = 0.01 kg-m2, B = 0.01 Nm/rads-1 and Tload = 5 Nm.
What is the torque profile (torque needed to be produced) ? 56

57. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile 10 25 45 60
t (ms)
speed
(rad/s)
100
0 < t <10 ms Te = 0.01(0) (0) + 5 Nm = 5 Nm
10ms < t <25 ms Te = 0.01(100/0.015) +0.01( t) + 5
= ( t) Nm
25ms < t< 45ms Te = 0.01(0) (100) + 5 = 6 Nm
45ms < t < 60ms Te = 0.01(-100/0.015) ( t) + 5
= – 66.67t 57

58. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
speed
(rad/s)
100
Speed profile 10 25 45 60
t (ms)
Torque
(Nm)
72.67
torque profile
71.67 6 5 10 25 45 60
t (ms)
-60.67
-61.67 58

59. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
Torque
(Nm) 70
J = kg-m2, B = 0.1 Nm/rads-1 and Tload = 5 Nm. 6 10 25 45 60
t (ms)
-65
For the same system and with the motor torque profile given above, what would be the speed profile? 59

60. Torque Equation: GraphicalTe
Speed
Forward
running
Forward
braking
Reverse
acc.
Reverse
running
Reverse
braking
Forward
acc.

61. Load Torque Load torque, TL, is complex, depending on applications.
In general:
TORQUE
TL = k
TL = kw
TL = kw2
SPEED

62. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Unavoidable power losses causes temperature increase
Insulation used in the windings are classified based on the temperature it can withstand.
Motors must be operated within the allowable maximum temperature
Sources of power losses (hence temperature increase):
- Conductor heat losses (i2R)
- Core losses – hysteresis and eddy current
- Friction losses – bearings, brush windage 62

63. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Steady-state stability 63

64. Figure 1.7. Typical load speed / torque, speed / power curves
Typical load torque / speed curves
Typical load torque / speed curves are shown on figure 1.7. They give a strong indication of the variety of torque / speed characteristics. Along such curves the mechanical power required from the motor varies with speed.
Figure 1.7. Typical load speed / torque, speed / power curves

65. 1.7. LOAD DYNAMICS AND STABILITY
(1.22)
(1.23)
where TS is the static friction torque (at zero speed); TC is Coulomb friction torque (constant with speed); TV is viscous friction torque (proportional to speed) and TW is windage friction (including the ventilator braking torque, proportional to speed squared):
(1.24)
(1.25)

66. Figure 1.12. Components of friction torque, Tfriction
Figure Mechanical characteristics:
a.) d.c. brush motor with separate excitation b.) induction motor
c.) synchronous motor

67. Various Motor Characteristics

68. Example 1.4. D.C. brush motor drive stability.
A permanent magnet d.c. brush motor with the torque speed curve: drives a d.c. generator which supplies a resistive load such that the generator torque / speed equation is Wr = 2TL. We calculate the speed and torque for the steady state point and find out if that point is stable.
Solution:
Let us first draw the motor and load (generator) torque speed curves on figure 1.14.
Figure D.C. brush motor load match

69. The steady state point, A, corresponds to constant speed and B = 0 in (1.27). Simply the motor torque counteracts the generator braking torque:
(1.37)
Using the two torque speed curves we find:
(1.38)
and thus:
(1.39)
and
(1.40)
The static stability is met if:
(1.41)
In our case from the two torque / speed curves:
(1.42)
and thus, as expected, point A represents a situation of static equilibrium.

70. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Electrical machines can be overloaded as long their temperature does not exceed the temperature limit
Accurate prediction of temperature distribution in machines is complex – hetrogeneous materials, complex geometrical shapes
Simplified assuming machine as homogeneous body
Ambient temperature, To p1
Thermal capacity, C (Ws/oC)
Surface A, (m2)
Surface temperature, T (oC) p2
Emitted heat power
(convection)
Input heat power
(losses) 70

71. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Power balance:
Heat transfer by convection:
, where  is the coefficient of heat transfer
Which gives:
With T(0) = 0 and p1 = ph = constant ,
, where 71

72. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Heating transient  t
Cooling transient t  72

73. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
The duration of overloading depends on the modes of operation:
Continuous duty
Load torque is constant over extended period multiple
Steady state temperature reached
Nominal output power chosen equals or exceeds continuous load
Continuous duty
Short time intermittent duty
Periodic intermittent duty
Losses due to continuous load t
p1n  73

74. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty
Operation considerably less than time constant, 
Motor allowed to cool before next cycle
Motor can be overloaded until maximum temperature reached 74

75. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty
p1s p1
p1n  t1 t 75

76. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty  t1 t 76

77. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Periodic intermittent duty
Load cycles are repeated periodically
Motors are not allowed to completely cooled
Fluctuations in temperature until steady state temperature is reached 77

78. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Periodic intermittent duty p1
heating
coolling t 78

79. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Periodic intermittent duty
Example of a simple case – p1 rectangular periodic pattern
pn = 100kW, nominal power
M = 800kg
= 0.92, nominal efficiency
T= 50oC, steady state temperature rise due to pn
Also,
If we assume motor is solid iron of specific heat cFE=0.48 kWs/kgoC, thermal capacity C is given by
C = cFE M = 0.48 (800) = 384 kWs/oC
Finally , thermal time constant = /180 = 35 minutes 79

80. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Periodic intermittent duty
Example of a simple case – p1 rectangular periodic pattern
For a duty cycle of 30% (period of 20 mins), heat losses of twice the nominal, 80

81. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque-speed quadrant of operation  1 2
T -ve
+ve
Pm -ve
T +ve
+ve
Pm +ve T 3 4
T -ve
-ve
Pm +ve
T +ve
-ve
Pm -ve 81

82. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
4-quadrant operation 
Direction of positive (forward) speed is arbitrary chosen Te m m Te
Direction of positive torque will produce positive (forward) speed
Quadrant 2
Forward braking
Quadrant 1
Forward motoring T
Quadrant 3
Reverse motoring
Quadrant 4
Reverse braking Te Te m m 82

83. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Ratings of converters and motors
Torque
Power limit for transient torque
Transient torque limit
Continuous torque limit
Power limit for continuous torque
Maximum
speed limit
Speed 83

84. 1.8. MULTIQUADRANT OPERATION
These possibilities are summarised in Table 1.1 and in figure 1.16.
Table 1.1.

85. 4Q OPERATION FB FM RB RM F: FORWARD R: REVERSE M : MOTORING B: BRAKING
SPEED w w Te Te FB FM II I
TORQUE RB
III IV w w RM Te Te

86. 4Q OPERATION: LIFT SYSTEM
Positive speed
Motor
Negative torque
Counterweight
Cage

87. 4Q OPERATION: LIFT SYSTEM
Convention:
Upward motion of the cage: Positive speed
Weight of the empty cage < Counterweight
Weight of the full-loaded cage > Counterweight
Principle:
What causes the motion?
Motor : motoring P =Tw = +ve
Load (counterweight) : braking P =Tw = -ve

88. 4Q OPERATION: LIFT SYSTEM
Speed
You are at 10th floor, calling
empty cage from gnd floor
You are at 10th floor, calling
fully-loaded cage from gnd floor FB FM
Torque RM RB
You are at gnd floor, calling
empty cage from 10th floor
You are at gnd floor, calling
Fully-loaded cage from 10th floor

89. DC MOTOR DRIVES Principle of operation Torque-speed characteristic
Methods of speed control
Armature voltage control
Variable voltage source
Phase-controlled Rectifier
Switch-mode converter (Chopper)
1Q-Converter
2Q-Converter
4Q-Converter

90. Figure 1.16. Electric drives with four quadrant operation