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ELECTRIC DRIVES
INTRODUCTION TO ELECTRIC DRIVES
MODULE 1
S. Samsudeen
Electrical Energy Conversion
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
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 1523 will be covering variable speed drives
Example on VSD application
motor pump
valve
Supply
Constant speed Variable Speed Drives
Power
In
Power loss
Mainly in valve
Power out
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Example on VSD application
motor pump
valve
Supply
motorPEC pump
Supply
Constant speed Variable Speed Drives
Power
In
Power loss
Power out
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Power loss
Mainly in valve
Power outPower
In
Power loss
Mainly in valve
Power out
motor pump
valve
Supply
motorPEC pump
Supply
Constant speed Variable Speed Drives
Example on VSD application
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Power
In
Power loss
Power
In
Power out
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Conventional electric drives (variable speed)
• Bulky
• Inefficient
• inflexible
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Modern electric drives (With power electronic converters)
• Small
• Efficient
• Flexible
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Modern electric drives
• Inter-disciplinary (PE, control system, machine
design, sensors)
• Several research area
• Expanding
Machine design
Speed sensorless
Machine Theory
Non-linear control
Real-time control
DSP application
PFC
Speed sensorless
Power electronic converters
Utility interface
Renewable energy
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
• Natural speed-torque characteristic is not compatible with load
requirements
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
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
• Electrical isolation between control circuit and power circuit is
needed:
• Malfuction in power circuit may damage control circuit
• Safety for the operator
• Avoid conduction of harmonic to control circuit
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Components in electric drives
Sensors
• Sensors (voltage, current, speed or torque) is normally
required for closed-loop operation or protection
• Electrical isolation between sensors and control circuit is
needed for the reasons previously explained
• The term ‘sensorless drives’ is normally referred to the drive
system where the speed is estimated rather than measured.
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives
Extracted from Boldea & Nasar
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
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
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Overview of AC and DC drives
After semiconductor devices were introduced (1950s)
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
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.
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
M
v
Fm
Ff
( )
dt
Mvd
FF fm =−
Newton’s law
Linear motion, constant M
• First order differential equation for speed
• Second order differential equation for displacement
( ) Ma
dt
xd
M
dt
vd
MFF 2
2
fm ===−
x
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
• First order differential equation for angular frequency (or velocity)
• Second order differential equation for angle (or position)
( )
2
2
m
le
dt
d
J
dt
d
JTT
θ
=
ω
=−
With constant J,
Rotational motion
- Normally is the case for electrical drives
( )
dt
Jd
TT m
le
ω
=−
θ
Te , ωm
Tl
J
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
dt
d
JTT m
le
ω
+=
For constant J,
( )
dt
d
J mω
Torque dynamic – present during speed transient
( )
dt
d mω Angular acceleration
Larger net torque and smaller J gives faster acceleration
0.19 0.2 0.21 0.22 0.23 0.24 0.25
-200
-100
0
100
200
speed(rad/s)
0.19 0.2 0.21 0.22 0.23 0.24 0.25
0
5
10
15
20
torque(Nm)
Elementary principles of mechanics
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
dt
d
JTT m
le
ω
+=
Elementary principles of mechanics
dt
d
JTT m
mlmem
ω
ω+ω=ω
dt
d
Jpp m
mLD
ω
ω+=⇒
Driving
power
Load
power
Change
in KE
• A step change in speed requires an infinite driving power
• Therefore ω is a continuous variable
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
dt
d
JTT m
le
ω
+=
Elementary principles of mechanics
dt
d
JTT m
mlmem
ω
ω+ω=ω
dt
d
Jpp m
mLD
ω
ω+=⇒
Integrating the equation with time and setting the initial speed ω(0) =
0, we obtain the following:
wD = pD dτ
0
t
∫ = pL dτ
0
t
∫ + ωm J
dωm
dτ0
t
∫ dτ
wD =wL +J ωm
0
ω
∫ dωm
wD = wL +
1
2
Jωm
2
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
A drive system that require fast acceleration must have
• small overall moment of inertia
• large motor torque capability
As the motor speed increases, the kinetic energy also increases.
During deceleration, the dynamic torque changes its sign and thus
helps motor to maintain the speed. This energy is extracted from the
stored kinetic energy:
J is purposely increased to do this job !
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Elementary principles of mechanics
( )
dt
vd
MFF le =−
Combination of rotational and translational motions
r r
ω
Te, ω
Tl
Fl Fe
v
M
Te = r(Fe), Tl = r(Fl), v =rω
dt
d
MrTT 2
le
ω
=−
r2
M - Equivalent moment inertia of the
linearly moving mass
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
ωm
ωm1
ωm2
n1
n2
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Motor
Te
Load 1,
Tl1
Load 2,
Tl2
J1
J2
ωm
ωm1
ωm2
n1
n2
Motor
Te
Jequ
Equivalent
Load , Tlequ
ωm
2
2
21equ JaJJ +=
Tlequ = Tl1 + a2Tl2
a2 = n1/n2=ω2/ω1
Elementary principles of mechanics – effect of gearing
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque-speed quadrant of operation
ω
T
12
3 4
T +ve
ω +ve
Pm +ve
T -ve
ω +ve
Pm -ve
T -ve
ω -ve
Pm +ve
T +ve
ω -ve
Pm -ve
• Quadrant of operation is
defined by the speed and
torque of the motor
• Most rotating electrical
machines can operate in 4
quadrants
• Not all converters can
operate in 4 quadrants
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque-speed quadrant of operation
ω
T
Te
ωm
Te
Te
Te
ωm
ωm
ωm
• Quadrant of operation is
defined by the speed and
torque of the motor
• Most rotating electrical
machines can operate in 4
quadrants
• Not all converters can
operate in 4 quadrants
Quadrant 1
Forward motoring
Quadrant 2
Forward braking
Quadrant 3
Reverse motoring
Quadrant 4
Reverse braking
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Motor steady state torque-speed characteristic (natural
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
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
SPEED
TORQUE
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
T~ C
Coulomb friction
T~ ω
Viscous friction
T~ ω2
Friction due to turbulent flow
α
TL
Te
Vehicle drive
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Constant torque, e.g. gravitational torque (active load)
SPEED
TORQUE
Gravitational torque
gM
FL
TL = rFL = r g M sin α
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Load steady state torque-speed characteristic
Hoist drive
Speed
Torque
Gravitational torque
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
TlTe
Steady state
speed
ωr
Torque
Speedωr2ωr3
ωr1
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
10 25 45 60 t (ms)
speed
(rad/s)
100
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) ?
Speed profile
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) + 0.01(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
le TB
dt
d
JT +ω+
ω
=
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
10 25 45 60
speed
(rad/s)
100
10 25 45 60
Torque
(Nm)
72.67
71.67
-60.67
-61.67
5
6
t (ms)
t (ms)
Speed profile
torque profile
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Torque and speed profile
10 25 45 60
Torque
(Nm)
70
-65
6
t (ms)
For the same system and with the motor torque profile
given above, what would be the speed profile?
J = 0.001 kg-m2, B = 0.1 Nm/rads-1
and Tload = 5 Nm.
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Ratings of converters and motors
Torque
Speed
Power limit for
continuous torque
Continuous
torque limit
Maximum
speed limit
Power limit for
transient torque
Transient
torque limit
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 (i2
R)
- Core losses – hysteresis and eddy current
- Friction losses – bearings, brush windage
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
p2
p1
Thermal capacity, C (Ws/o
C)
Surface A, (m2
)
Surface temperature, T (o
C)Input heat power
(losses)
Emitted heat power
(convection)
Ambient temperature, To
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Power balance:
21 pp
dt
dT
C −=
Heat transfer by convection:
)TT(Ap o2 −α=
C
p
T
C
A
dt
Td 1
=∆
α
+
∆
Which gives:
( )τ−
−
α
=∆ /th
e1
A
p
T
A
C
α
=τ, where
With ∆T(0) = 0 and p1 = ph = constant ,
, where α is the coefficient of heat transfer
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
tτ
T∆
t
τ
τ−
⋅∆=∆ /t
e)0(TT
T∆
( )τ−
−
α
=∆ /th
e1
A
p
T
Heating transient
Cooling transient
A
ph
α
)0(T∆
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
The duration of overloading depends on the modes of operation:
Continuous duty
Short time intermittent duty
Periodic intermittent duty
Continuous duty
Load torque is constant over extended period multiple
Steady state temperature reached
Nominal output power chosen equals or exceeds continuous load
T∆
t
A
p n1
α
τ
p1n
Losses due to continuous load
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
t1
τ
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty
A
p s1
α
maxT∆
A
p n1
α
t
T∆
p1
p1n
p1s
t1
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Short time intermittent duty
τ t
T∆
( )τ−
−
α
=∆ /ts1
e1
A
p
T
maxT∆
A
p n1
α
( )τ−
−
α
=
α
/ts1n1 1
e1
A
p
A
p ( )τ−
−≥ /t
s1n1
1
e1pp
1
/t
n1
s1
te1
1
p
p
1
τ
≈
−
≤ τ−
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
INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
Thermal considerations
Periodic intermittent duty
p1
t
heating coolling
coolling
coolling
heating
heating
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∞= 50o
C, steady state temperature rise due to pn
kW91
1
pp n1 =





−
η
= Also, C/W180
50
9000
T
p
A o1
==
∆
=α
∞
If we assume motor is solid iron of specific heat cFE=0.48 kWs/kgo
C,
thermal capacity C is given by
C = cFE M = 0.48 (800) = 384 kWs/o
C
Finally τ, thermal time constant = 384000/180 = 35 minutes
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,
0 0.5 1 1.5 2 2.5
x 10
4
0
5
10
15
20
25
30
35

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Electric drives

  • 1. ELECTRIC DRIVES INTRODUCTION TO ELECTRIC DRIVES MODULE 1 S. Samsudeen Electrical Energy Conversion
  • 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 1523 will be covering variable speed drives
  • 4. Example on VSD application motor pump valve Supply Constant speed Variable Speed Drives Power In Power loss Mainly in valve Power out INTRODUCTION TO ELECTRIC DRIVES - MODULE 1
  • 5. Example on VSD application motor pump valve Supply motorPEC pump Supply Constant speed Variable Speed Drives Power In Power loss Power out INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Power loss Mainly in valve Power outPower In
  • 6. Power loss Mainly in valve Power out motor pump valve Supply motorPEC pump Supply Constant speed Variable Speed Drives Example on VSD application INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Power In Power loss Power In Power out
  • 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 • Inter-disciplinary (PE, control system, machine design, sensors) • Several research area • Expanding Machine design Speed sensorless Machine Theory Non-linear control Real-time control DSP application PFC Speed sensorless Power electronic converters Utility interface Renewable energy
  • 10. 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 • Natural speed-torque characteristic is not compatible with load requirements 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
  • 11. 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 • Electrical isolation between control circuit and power circuit is needed: • Malfuction in power circuit may damage control circuit • Safety for the operator • Avoid conduction of harmonic to control circuit
  • 12. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Components in electric drives Sensors • Sensors (voltage, current, speed or torque) is normally required for closed-loop operation or protection • Electrical isolation between sensors and control circuit is needed for the reasons previously explained • The term ‘sensorless drives’ is normally referred to the drive system where the speed is estimated rather than measured.
  • 13. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Overview of AC and DC drives Extracted from Boldea & Nasar
  • 14. 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
  • 15. 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
  • 16. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Overview of AC and DC drives After semiconductor devices were introduced (1950s)
  • 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 M v Fm Ff ( ) dt Mvd FF fm =− Newton’s law Linear motion, constant M • First order differential equation for speed • Second order differential equation for displacement ( ) Ma dt xd M dt vd MFF 2 2 fm ===− x
  • 20. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Elementary principles of mechanics • First order differential equation for angular frequency (or velocity) • Second order differential equation for angle (or position) ( ) 2 2 m le dt d J dt d JTT θ = ω =− With constant J, Rotational motion - Normally is the case for electrical drives ( ) dt Jd TT m le ω =− θ Te , ωm Tl J
  • 21. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Elementary principles of mechanics
  • 22. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 dt d JTT m le ω += For constant J, ( ) dt d J mω Torque dynamic – present during speed transient ( ) dt d mω Angular acceleration Larger net torque and smaller J gives faster acceleration 0.19 0.2 0.21 0.22 0.23 0.24 0.25 -200 -100 0 100 200 speed(rad/s) 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0 5 10 15 20 torque(Nm) Elementary principles of mechanics
  • 23. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 dt d JTT m le ω += Elementary principles of mechanics dt d JTT m mlmem ω ω+ω=ω dt d Jpp m mLD ω ω+=⇒ Driving power Load power Change in KE • A step change in speed requires an infinite driving power • Therefore ω is a continuous variable
  • 24. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 dt d JTT m le ω += Elementary principles of mechanics dt d JTT m mlmem ω ω+ω=ω dt d Jpp m mLD ω ω+=⇒ Integrating the equation with time and setting the initial speed ω(0) = 0, we obtain the following: wD = pD dτ 0 t ∫ = pL dτ 0 t ∫ + ωm J dωm dτ0 t ∫ dτ wD =wL +J ωm 0 ω ∫ dωm wD = wL + 1 2 Jωm 2
  • 25. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Elementary principles of mechanics A drive system that require fast acceleration must have • small overall moment of inertia • large motor torque capability As the motor speed increases, the kinetic energy also increases. During deceleration, the dynamic torque changes its sign and thus helps motor to maintain the speed. This energy is extracted from the stored kinetic energy: J is purposely increased to do this job !
  • 26. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Elementary principles of mechanics ( ) dt vd MFF le =− Combination of rotational and translational motions r r ω Te, ω Tl Fl Fe v M Te = r(Fe), Tl = r(Fl), v =rω dt d MrTT 2 le ω =− r2 M - Equivalent moment inertia of the linearly moving mass
  • 27. 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 ωm ωm1 ωm2 n1 n2
  • 28. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Motor Te Load 1, Tl1 Load 2, Tl2 J1 J2 ωm ωm1 ωm2 n1 n2 Motor Te Jequ Equivalent Load , Tlequ ωm 2 2 21equ JaJJ += Tlequ = Tl1 + a2Tl2 a2 = n1/n2=ω2/ω1 Elementary principles of mechanics – effect of gearing
  • 29. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Torque-speed quadrant of operation ω T 12 3 4 T +ve ω +ve Pm +ve T -ve ω +ve Pm -ve T -ve ω -ve Pm +ve T +ve ω -ve Pm -ve • Quadrant of operation is defined by the speed and torque of the motor • Most rotating electrical machines can operate in 4 quadrants • Not all converters can operate in 4 quadrants
  • 30. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Torque-speed quadrant of operation ω T Te ωm Te Te Te ωm ωm ωm • Quadrant of operation is defined by the speed and torque of the motor • Most rotating electrical machines can operate in 4 quadrants • Not all converters can operate in 4 quadrants Quadrant 1 Forward motoring Quadrant 2 Forward braking Quadrant 3 Reverse motoring Quadrant 4 Reverse braking
  • 31. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Motor steady state torque-speed characteristic (natural 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
  • 32. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Load steady state torque-speed characteristic SPEED TORQUE 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 T~ C Coulomb friction T~ ω Viscous friction T~ ω2 Friction due to turbulent flow
  • 33. α TL Te Vehicle drive INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Load steady state torque-speed characteristic Constant torque, e.g. gravitational torque (active load) SPEED TORQUE Gravitational torque gM FL TL = rFL = r g M sin α
  • 34. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Load steady state torque-speed characteristic Hoist drive Speed Torque Gravitational torque
  • 35. 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 TlTe Steady state speed ωr Torque Speedωr2ωr3 ωr1
  • 36. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Torque and speed profile 10 25 45 60 t (ms) speed (rad/s) 100 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) ? Speed profile
  • 37. 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) + 0.01(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 le TB dt d JT +ω+ ω =
  • 38. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Torque and speed profile 10 25 45 60 speed (rad/s) 100 10 25 45 60 Torque (Nm) 72.67 71.67 -60.67 -61.67 5 6 t (ms) t (ms) Speed profile torque profile
  • 39. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Torque and speed profile 10 25 45 60 Torque (Nm) 70 -65 6 t (ms) For the same system and with the motor torque profile given above, what would be the speed profile? J = 0.001 kg-m2, B = 0.1 Nm/rads-1 and Tload = 5 Nm.
  • 40. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Ratings of converters and motors Torque Speed Power limit for continuous torque Continuous torque limit Maximum speed limit Power limit for transient torque Transient torque limit
  • 41. 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 (i2 R) - Core losses – hysteresis and eddy current - Friction losses – bearings, brush windage
  • 42. 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 p2 p1 Thermal capacity, C (Ws/o C) Surface A, (m2 ) Surface temperature, T (o C)Input heat power (losses) Emitted heat power (convection) Ambient temperature, To
  • 43. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Thermal considerations Power balance: 21 pp dt dT C −= Heat transfer by convection: )TT(Ap o2 −α= C p T C A dt Td 1 =∆ α + ∆ Which gives: ( )τ− − α =∆ /th e1 A p T A C α =τ, where With ∆T(0) = 0 and p1 = ph = constant , , where α is the coefficient of heat transfer
  • 44. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Thermal considerations tτ T∆ t τ τ− ⋅∆=∆ /t e)0(TT T∆ ( )τ− − α =∆ /th e1 A p T Heating transient Cooling transient A ph α )0(T∆
  • 45. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Thermal considerations The duration of overloading depends on the modes of operation: Continuous duty Short time intermittent duty Periodic intermittent duty Continuous duty Load torque is constant over extended period multiple Steady state temperature reached Nominal output power chosen equals or exceeds continuous load T∆ t A p n1 α τ p1n Losses due to continuous load
  • 46. 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
  • 47. t1 τ INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Thermal considerations Short time intermittent duty A p s1 α maxT∆ A p n1 α t T∆ p1 p1n p1s
  • 48. t1 INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Thermal considerations Short time intermittent duty τ t T∆ ( )τ− − α =∆ /ts1 e1 A p T maxT∆ A p n1 α ( )τ− − α = α /ts1n1 1 e1 A p A p ( )τ− −≥ /t s1n1 1 e1pp 1 /t n1 s1 te1 1 p p 1 τ ≈ − ≤ τ−
  • 49. 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
  • 50. INTRODUCTION TO ELECTRIC DRIVES - MODULE 1 Thermal considerations Periodic intermittent duty p1 t heating coolling coolling coolling heating heating
  • 51. 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∞= 50o C, steady state temperature rise due to pn kW91 1 pp n1 =      − η = Also, C/W180 50 9000 T p A o1 == ∆ =α ∞ If we assume motor is solid iron of specific heat cFE=0.48 kWs/kgo C, thermal capacity C is given by C = cFE M = 0.48 (800) = 384 kWs/o C Finally τ, thermal time constant = 384000/180 = 35 minutes
  • 52. 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, 0 0.5 1 1.5 2 2.5 x 10 4 0 5 10 15 20 25 30 35