introduction to electric drives module...

ELECTRIC DRIVES

INTRODUCTION TO ELECTRIC DRIVES

MODULE 1

Dr. Nik Rumzi Nik Idris

Dept. of Energy Conversion, UTM

2013

INTRODUCTION TO ELECTRIC DRIVES - MODULE 1

Electrical Drives

Drivesare systems employed for motion control

Require prime movers

Drives that employ electric motors as prime movers are known as Electrical Drives


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 lossMainly in valve

Power out



motor pump

valve

Supply

motorPEC pump

Supply


Power

In

Power loss

Power out



Power outPower

In


Power out

motor pump

valve

Supply

motorPEC pump

Supply




Power

In

Power loss

Power

In

Power out


Conventional electric drives (variable speed)

• Bulky

• Inefficient

• inflexible


Modern electric drives (With power electronic converters)

• Small

• Efficient

• Flexible


Modern electric drives

• Inter-disciplinary

• Several research area

• Expanding

Machine designSpeed sensorlessMachine Theory

Non-linear controlReal-time controlDSP application

PFCSpeed sensorless

Power electronic converters

Utility interfaceRenewable energy


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



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



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.


Overview of AC and DC drives

Extracted from Boldea & Nasar



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



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



After semiconductor devices were introduced (1950s)



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


Classification of IM drives (Buja, Kamierkowski, ““““Direct torque control of PWM inverter-fed AC motors - a survey””””,

IEEE Transactions on Industrial Electronics, 2004.


Elementary principles of mechanics

M

v

Fm

Ff

( )dt

MvdFF fm =−

Newton’s law

Linear motion, constant M

• First order differential equation for speed• Second order differential equation for displacement

( )Ma

dt

xdM

dt

vdMFF

2

2

fm ===−

x



• First order differential equation for angular frequency (or velocity)• Second order differential equation for angle (or position)

( )2

2

m

ledt

dJ

dt

dJTT

θ=

ω=−

With constant J,

Rotational motion

- Normally is the case for electrical drives

( )dt

JdTT mle

ω=−

θ

Te , ωm

T l

J


dt

dJTT m

le

ω+=

For constant J,

( )dt

dJ m

ωTorque dynamic – present during speed transient

( )dt

dm

ω 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)




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 !



( )dt

vdMFF le =−

Combination of rotational and translational motions

r r

ωTe, ω

T l

Fl Fe

v

M

Te = r(Fe), T l = r(Fl), v =rω

dt

dMrTT 2

le

ω=−

r2M - Equivalent moment inertia of the linearly moving mass


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

TeLoad 1,

Tl1

Load 2,

Tl2J1

J2

ωmωm1

ωm2

n1

n2


Motor

Te

Load 1,

Tl1

Load 2,

Tl2J1

J2

ωmωm1

ωm2

n1

n2

Motor

Te

Jequ

Equivalent

Load , Tlequ

ωm2

2

21equ JaJJ +=

Tlequ = Tl1 + a2Tl2

a2 = n1/n2=ω2/ω1

Elementary principles of mechanics – effect of gearing


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


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

αααα

TL

Te

Vehicle drive



Constant torque, e.g. gravitational torque (active load)

SPEED

TORQUE

Gravitational torque

gM

FL

TL = rFL = r g M sin αααα



Hoist drive

Speed

Torque

Gravitational torque


Load and motor steady state torque

At constant speed, Te= T l

Steady state speed is at point of intersection between Te and T l of the

steady state torque characteristics

T lTe

Steady state speed

ωr

Torque

Speedωr2ωr3 ωr1


Torque and speed profile

10 25 45 60 t (ms)

speed (rad/s)

100

The system is described by: Te – T load = 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



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 TBdt

dJT +ω+ω

=



10 25 45 60

speed (rad/s)

100

10 25 45 60

Torque (Nm)

72.67

71.67

-60.67

-61.67

56

t (ms)

t (ms)

Speed profile

torque 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.


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



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 homogeneousbody

p2p1

Thermal capacity, C (Ws/oC)Surface A, (m2)Surface temperature, T (oC)Input heat power

(losses)

Emitted heat power(convection)

Ambient temperature, To



Power balance:

21 ppdt

dTC −=

Heat transfer by convection:

)TT(Ap o2 −α=

C

pT

C

A

dt

Td 1=∆α

+∆

Which gives:

( )τ−−α

=∆ /th e1A

pT

A

C

α=τ, where

With ∆T(0) = 0 and p1 = ph = constant ,

, where α is the coefficient of heat transfer



tτ

T∆

tτ

τ−⋅∆=∆ /te)0(TT

T∆

( )τ−−α

=∆ /th e1A

pT

Heating transient

Cooling transient

A

ph

α

)0(T∆



The duration of overloading depends on the modes of operation:

Continuous dutyShort 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



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

t1τ




A

p s1

α

maxT∆ A

p n1

α

t

T∆

p1

p1n

p1s

t1




τ t

T∆

( )τ−−α

=∆ /ts1 e1A

pT

maxT∆A

p n1

α

( )τ−−α

=α

/ts1n1 1e1A

p

A

p ( )τ−−≥ /t

s1n11e1pp

1

/t

n1

s1

te1

1

p

p1

τ≈

−≤ τ−




Load cycles are repeated periodically

Motors are not allowed to completely cooled

Fluctuations in temperature until steady state temperature is reached




p1

t

heating coollingcoolling

coolling

heating

heating




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

kW911

pp n1 =

−

η= Also, C/W180

50

9000

T

pA o1 ==

∆=α

∞

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




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 104

0

5

10

15

20

25

30

35


Torque-speed quadrant of operation

ω

T

12

3 4

T +veω +vePm +ve

T -veω +vePm -ve

T -veω -vePm +ve

T +veω -vePm -ve


4-quadrant operation

ωm

Te

Te

ωm

Te

ωm

Te

ωm

ω

T

• Direction of positive (forward) speed is arbitrary chosen

• Direction of positive torque will produce positive (forward) speed

Quadrant 1Forward motoring

Quadrant 2Forward braking

Quadrant 3Reverse motoring

Quadrant 4Reverse braking


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


Steady-state stability

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