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ELECTRIC DRIVES

INTRODUCTION TO ELECTRIC DRIVES

MODULE 1

Dr. Nik Rumzi Nik Idris

Dept. of Energy Conversion, UTM

2006

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INTRODUCTION TO ELECTRIC DRIVES - MODULE 1

Electrical Drives

Drives are systems employed formotion control

Require prime movers

Drives that employ electric motors as

prime movers are known as Electrical Drives

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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

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Example on VSD application

motor pump

valve

Supply

Constant speed Variable Speed Drives

Power

In

Power lossMainly in valve

Power out


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motor pump

valve

Supply

motorPEC pump

Supply


Power

In

Power loss

Power out



Power outPower

In

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Power out

motor pump

valve

Supply

motorPEC pump

Supply




Power

In

Power loss

Power

In

Power out

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Conventional electric drives (variable speed)

Bulky

Inefficient

inflexible

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Modern electric drives (With power electronic converters)

Small

Efficient

Flexible

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Modern electric drives

Inter-disciplinary

Several research area

Expanding

Machine designSpeed sensorlessMachine Theory

Non-linear controlReal-time controlDSP application

PFCSpeed sensorlessPower electronic converters

Utility interface

Renewable energy

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Components in electric drives

e.g. Single drive - sensorless vector control from Hitachi

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e.g. Multidrives system from ABB

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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

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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 controllers

DSP/microprocessor flexible, lower bandwidth - DSPs perform

faster operation than microprocessors (multiplication in single

cycle), can perform complex estimations

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Overview of AC and DC drives

Extracted from Boldea & Nasar

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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

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Before semiconductor devices were introduced (

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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

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Classification of IM drives (Buja, Kamierkowski, Direct torque control of PWM inverter-fed AC motors - a survey,IEEE Transactions on Industrial Electronics, 2004.

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Elementary principles of mechanics

M

v

Fm

Ff

dt

MvdFF fm !

Newtons 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

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First order differential equation for angular frequency (or velocity)

Second order differential equation for angle (or position)

2

2

m

le dt

dJ

dt

dJTT

U

!

[

!

With constant J,

Rotational motion

- Normally is the case for electrical drives

dt

JdTT mle

[!

U

Te , [m

Tl

J

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dtdJTT m

le[!For constant J,

dt

dJ

m[

Torque dynamic present during speed transient

dt

dm

[Angular acceleration (speed)

The larger the net torque, the faster the acceleration is.

0.19 0.2 0. 21 0. 22 0. 23 0. 24 0. 25-200

-10 0

0

10 0

200

s

peed

(rad/s

)

0.19 0.2 0. 21 0. 22 0. 23 0. 24 0. 25

0

5

10

15

20

torque

(Nm)


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dt

vdMFF

le!

Combination of rotational and translational motions

r r

[Te, [

Tl

Fl Fe

v

M

Te = r(Fe), Tl = r(Fl), v =r[

dt

dMrTT 2

le

[!

r2M - Equivalent moment inertia of the

linearly moving mass

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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

MotorTe

Load 1,

Tl1

Load 2,

Tl2

J1

J2

[m

[m1

[m2

n1

n2

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Motor

Te

Load 1,

Tl1

Load 2,

Tl2

J1

J2

[m

[m1

[m2

n1

n2

Motor

Te

Jequ

Equivalent

Load , Tlequ

[m2

2

21equ JaJJ !

Tlequ = Tl1 + a2Tl2

a2 = n1/n2

Elementary principles of mechanics effect of gearing

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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

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Load steady state torque-speed characteristic

SPEED

TORQUE

Frictional torque (passive load) Exist in all motor-load drivesystem simultaneously

In most cases, only one or two

are dominating

Exists when there is motion

T~ C

Coulomb friction

T~ [

Viscous friction

T~ [2

Friction due to turbulent flow

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E

TL

Te

Vehicle drive



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

SPEED

TORQUE

Gravitational torque

gM

FL

TL = rFL = r g M sin E

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Hoist drive

Speed

Torque

Gravitational torque

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Load and motor steady state torque

At constant speed, Te= TlSteady state speed is at point of intersection between Te and Tl of the

steady state torque characteristics

TlTe

Steady state

speed

[r

Torque

Speed[r2[r3 [r1

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Torque and speed profile

10 25 45 60 t (ms)

speed

(rad/s)

100

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) ?

Speed profile

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10 25 45 60 t (ms)

speed

(rad/s)

100

0 < t

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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

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10 25 45 60

Torque

(Nm)

70

-65

6

t (ms)

For the same system and with the motor torque profilegiven above, what would be the speed profile?

J = 0.001 kg-m2, B = 0.1 Nm/rads-1and Tload = 5 Nm.

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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

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Electrical machines can be overloaded as long their temperaturedoes 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

p2p

1 Thermal capacity, C (Ws/oC)Surface A, (m2)

Surface temperature, T (oC)Input heat power

(losses)

Emitted heat power

(convection)

Ambient temperature, To

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Power balance:

21 ppdt

dTC !

Heat transfer by convection:

)TT(Ap o2 E!

C

pT

C

A

dt

Td 1!(E

(

Which gives:

XE

!( /th e1A

pT

A

C

E!X, where

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

, whereE

is the coefficient of heat transfer

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tX

T(

tX

X(!( /te)0(TT

T(

XE!(

/th e1A

pT

Heating transient

Cooling transient

A

ph

E

)0(T(

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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 n1E

X

p1n

Losses due to continuous load

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Operation considerably less than time constant, X

Motor allowed to cool before next cycle

Motor can be overloaded until maximum temperature reached

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t1X




Ap s1E

maxT( A

p n1

E

t

T(

p1

p1n

p1s

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t1




X t

T(

XE

!( /ts1 e1A

pT

maxT(A

p n1

E

X

E!E/ts1n1 1

e1A

p

A

p

X

u

/t

s1n1

1

e1pp1

/tn1

s1

te1

1

p

p

1

X}

e

X

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Load cycles are repeated periodically

Motors are not allowed to completely cooled

Fluctuations in temperature until steady state temperature is reached

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p1

t

heating coollingcoolling

coolling

heating

heating

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Example of a simple case p1 rectangular periodic pattern

pn = 100kW, nominal power

M = 800kg

L= 0.92, nominal efficiency

(Tg= 50oC, steady state temperature rise due to pn

kW911

pp n1 !

L! Also, C/W180

50

9000

T

pA o1 !!

(!E

g

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 X, thermal time constant = 384000/180 = 35 minutes

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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

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Torque-speed quadrant of operation

[

T12

3 4

T +ve

[ +vePm +ve

T -ve

[ +vePm -ve

T -ve

[ -vePm +ve

T +ve

[ -vePm -ve

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Ratings of converters and motors

Torque

Speed

Powerlimit for

continuoustorque

Continuous

torquelimit

Maximum

speed limit

Powerlimit for

transienttorque

Transient

torquelimit

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Steady-state stability

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