2. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Electrical Drives
Drives are systems employed for motion control
Require prime movers
Drives that employ electric motors as
prime movers are known as Electrical Drives
3. 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)
• MEP 1522 will be covering variable speed drives
4. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
Constant speed Variable Speed Drives
valve
Supply
motor pump
Power Power out
In
Power loss
Mainly in valve
5. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
Constant speed Variable Speed Drives
valve
Supply Supply
motor pump motor
PEC pump
Power Power out
In Power Power out
In
Power loss
Power loss
Mainly in valve
6. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
Constant speed Variable Speed Drives
valve
Supply Supply
motor pump motor
PEC pump
Power Power out
In Power Power out
In
Power loss
Power loss
Mainly in valve
7. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Conventional electric drives (variable speed)
• Bulky
• Inefficient
• inflexible
8. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Modern electric drives (With power electronic converters)
• Small
• Efficient
• Flexible
9. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Modern electric drives
Machine design
Utility interface
Speed sensorless
Renewable energy
Machine Theory
Non-linear control
Real-time control
DSP application
PFC
Speed sensorless
Power electronic converters
• Inter-disciplinary
• Several research area
• Expanding
10. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
e.g. Single drive - sensorless vector control from Hitachi
11. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
e.g. Multidrives system from ABB
12. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
Motors
• DC motors - permanent magnet – wound field
• AC motors – induction, synchronous (IPMSM, SMPSM),
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
13. 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
14. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives
Extracted from Boldea & Nasar
15. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives
DC motors: Regular maintenance, heavy, expensive, speed limit
Easy control, decouple control of torque and flux
AC motors: Less maintenance, light, less expensive, high speed
Coupling between torque and flux – variable
spatial angle between rotor and stator flux
16. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives
Before semiconductor devices were introduced (<1950)
• AC motors for fixed speed applications
• DC motors for variable speed applications
After semiconductor devices were introduced (1950s)
• Variable frequency sources available – AC motors in variable
speed applications
• Coupling between flux and torque control
• Application limited to medium performance applications –
fans, blowers, compressors – scalar control
• High performance applications dominated by DC motors –
tractions, elevators, servos, etc
17. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives
After vector control drives were introduced (1980s)
• AC motors used in high performance applications – elevators,
tractions, servos
• AC motors favorable than DC motors – however control is
complex hence expensive
• Cost of microprocessor/semiconductors decreasing –predicted
30 years ago AC motors would take over DC motors
18. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Classification of IM drives (Buja, Kamierkowski, “Direct torque control of PWM inverter-fed AC motors - a survey”,
IEEE Transactions on Industrial Electronics, 2004.
19. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
v
x Newton’s law
Fm d( Mv )
M Fm − Ff =
Ff dt
Linear motion, constant M
d( v ) d2 x
Fm − Ff = M = M 2 = Ma
dt dt
• First order differential equation for speed
• Second order differential equation for displacement
20. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
θ Rotational motion
- Normally is the case for electrical drives
Tl
d( Jωm )
Te − Tl =
Te , ω m dt
J
With constant J,
d( ωm ) d2θ
Te − Tl = J =J 2
dt dt
• First order differential equation for angular frequency (or velocity)
• Second order differential equation for angle (or position)
21. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
For constant J, dωm
Te = Tl + J
dt
d( ωm )
J Torque dynamic – present during speed transient
dt
d( ωm ) Angular acceleration (speed)
dt
The larger the net torque, the faster the acceleration is.
200
100
speed (rad/s)
0
-100
-200
0.19 0.2 0.21 0.22 0.23 0.24 0.25
20
15
torque (Nm)
10
5
0
0.19 0.2 0.21 0.22 0.23 0.24 0.25
22. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
Combination of rotational and translational motions
Fl Fe
Te, ω
ω
r M r
Tl
v
d( v )
Fe − Fl = M Te = r(Fe), Tl = r(Fl), v =rω
dt
dω
Te − Tl = r 2M
dt
r2M - Equivalent moment inertia of the
linearly moving mass
23. 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
ωm ωm1
Motor Load 1, n1
Te Tl1
J2
ωm2
n2 Load 2,
J1 Tl2
24. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics – effect of gearing
ωm ωm1
Motor Load 1, n1
Te Tl1
J2
ωm2
n2 Load 2,
J1
Tl2
J equ = J1 + a 2 J 2
2
ωm
Motor Equivalent
Te Load , Tlequ
Tlequ = Tl1 + a2Tl2
Jequ
a2 = n1/n2
25. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Motor steady state torque-speed characteristic
SPEED
Synchronous mch
Induction mch
Separately / shunt DC mch
Series DC
TORQUE
By using power electronic converters, the motor characteristic
can be change at will
26. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Frictional torque (passive load) • Exist in all motor-load drive
SPEED
T~ C system simultaneously
T~ ω2
• In most cases, only one or two
T~ ω are dominating
• Exists when there is motion
TORQUE
Coulomb friction
Viscous friction
Friction due to turbulent flow
27. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Constant torque, e.g. gravitational torque (active load)
SPEED Gravitational torque
Vehicle drive
Te
TORQUE
TL
gM
α
FL
TL = rFL = r g M sin α
28. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Hoist drive
Speed
Torque
Gravitational torque
29. 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
Torque Te Tl
Steady state
speed
ωr3 ωr
ωr1 ωr2 Speed
30. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
speed Speed profile
(rad/s)
100
10 25 45 60 t (ms)
The system is described by: Te – Tload = J(dω/dt) + Bω
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) ?
31. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
speed
(rad/s)
dω
100
Te = J + Bω + Tl
dt
10 25 45 60 t (ms)
0 < t <10 ms Te = 0.01(0) + 0.01(0) + 5 Nm = 5 Nm
10ms < t <25 ms Te = 0.01(100/0.015) +0.01(-66.67 + 6666.67t) + 5
= (71 + 66.67t) Nm
25ms < t< 45ms Te = 0.01(0) + 0.01(100) + 5 = 6 Nm
45ms < t < 60ms Te = 0.01(-100/0.015) + 0.01(400 -6666.67t) + 5
= -57.67 – 66.67t
32. 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
33. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
Torque
(Nm)
70 J = 0.001 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?
34. 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
35. 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 p2
Thermal capacity, C (Ws/oC)
Input heat power
Surface A, (m2) Emitted heat power
Surface temperature, T (oC)
(losses) (convection)
36. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Power balance:
dT
C = p1 − p 2
dt
Heat transfer by convection:
p 2 = αA (T − To ) , where α is the coefficient of heat transfer
Which gives:
d∆T Aα p
+ ∆T = 1
dt C C
With ∆T(0) = 0 and p1 = ph = constant ,
∆T =
ph
αA
(
1 − e −t / τ ) , where τ =
C
αA
37. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
ph
∆T αA ∆T =
ph
αA
(1 − e −t / τ )
Heating transient
t
τ
∆T
∆T = ∆T(0) ⋅ e − t / τ
∆T(0)
Cooling transient
τ t
38. 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
Continuous duty
Short time intermittent duty
Steady state temperature reached
Periodic intermittent duty
Nominal output power chosen equals or exceeds continuous load
p1n
Losses due to continuous load
∆T αA
p1n
t
τ
39. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty
Operation considerably less than time constant, τ
Motor allowed to cool before next cycle
Motor can be overloaded until maximum temperature reached
40. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty p1s
p1
p1n
∆T p1s
αA
p1n
αA
∆Tmax
τ t
t1
41. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
p1s p p1 τ
Short time intermittent duty ≤ 1n ≥=p 1(s1 −1eτ e −t )/ τ )
p ( ≈t
− t 1 /−
− t1 / τ
1
∆T p1n αA − αA
1 e 1n 1s
1
∆T =
p1s
αA
(1 − e −t / τ )
p1n
αA
∆Tmax
τ t
t1
42. 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
43. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Periodic intermittent duty
p1
heating coolling
heating coolling
heating coolling
t
44. 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
η= 0.92, nominal efficiency
∆T∞= 50oC, steady state temperature rise due to pn
1 p1 9000
p1 = p n − 1 = 9kW
η Also, αA = = = 180 W / o C
∆T∞ 50
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 τ, thermal time constant = 384000/180 = 35 minutes
45. 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,
35
30
25
20
15
10
5
0
0 0.5 1 1.5 2 2.5
4
x10
46. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque-speed quadrant of operation
ω
2 1
T -ve T +ve
ω +ve ω +ve
Pm -ve Pm +ve
T
3 4
T -ve T +ve
ω -ve ω -ve
Pm +ve Pm -ve
47. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
4-quadrant operation
ω Te
• Direction of positive (forward)
speed is arbitrary chosen
ωm ωm
Te • Direction of positive torque will
produce positive (forward) speed
Quadrant 2 Quadrant 1
Forward braking Forward motoring
T
Quadrant 3 Quadrant 4
Reverse motoring Reverse braking Te
Te
ωm ωm
48. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Ratings of converters and motors
Torque
Transient Power limit for
torque limit transient torque
Continuous
torque limit Power limit for
continuous torque
Maximum
speed limit
Speed