Electric Drivesnew

8/2/2019 Electric Drivesnew

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ELECTRIC DRIVES

INTRODUCTION TO ELECTRIC DRIVES


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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


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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 design

Speed sensorless

Machine Theory

Non-linear control

Real-time control

DSP application

PFC

Speed sensorless

Power electronic converters

Utility interface

Renewable energy


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Components in electric drives

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 efficientFlexibleCompactAC-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 performfaster 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 predicted30 years ago AC motors would take over DC motors


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Elementary principles of mechanics

M

v

Fm

FfFmFf=

d (Mv )dt

Newtons law

Linear motion, constant M

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

FmF

f=M

d (v )dt

=Md2x

dt2=Ma

x


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

TeT

l=J

d (m )

dt

=Jd2

dt2

With constant J,

Rotational motion

- Normally is the case for electrical drives

TeT

l=d (Jm )dt

Te , m

Tl

J


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Te=Tl+J

dm

dt

For constant J,

Jd (m)dt

Torque dynamic present during speed transient

d (m )dt

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

-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)



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FeF

l=M

d ( v )dt

Combination of rotational and translational motions

r r

Te,

Tl

Fl Fe

v

M

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

TeT

l=r

2M

d

dt

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

mm1

m2

n1

n2


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Motor

Te

Load 1,

Tl1

Load 2,

Tl2J1

J2

m

m1

m2

n1

n2

Motor

Te

Jequ

Equivalent

Load , Tlequ

mJequ

=J1

+a2

2J2

Tlequ = Tl1 + a2Tl2

a2 = n1/n2

Elementary principles of mechanics effect of gearing


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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


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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~

Viscous friction

T~ 2

Friction due to turbulent flow


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TL

Te

Vehicle drive


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

SPEED

TORQUE

Gravitational torque

gM

FL

TL = rFL = r g M sin


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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

Speedr2r3 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-1

and 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

p2

p1Thermal 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:

CdTdt

=p1p2

Heat transfer by convection:

p2=A (TTo)

dT

dt+A

CT=

p1

C

Which gives:

T=p

h

A(1et/) =

C

A, where

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

, where is the coefficient of heat transfer


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t

T

t

T=T(0 )et/T

T=p

h

A

(1et/)

Heating transient

Cooling transient

ph

A

T(0 )


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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

p1n

A

p1n

Losses due to continuous load


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Operation considerably less than time constant,

Motor allowed to cool before next cycle

Motor can be overloaded until maximum temperature reached


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t1



p1s

A

Tmax

p1n

A

t

T

p1

p1n

p1s


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t1



t

T

T=p

1s

A(1et/)

Tmax

p1n

A

p1n

A

=p

1s

A

(1et1/)p1np1s (1et1/)

p1s

p1n

1

1et

1/

t1


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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 coolling coolling

coolling

heating

heating


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Example of a simple case p1 rectangular periodic pattern

pn = 100kW, nominal power

M = 800kg

= 0.92, nominal efficiency

T= 50o

C, steady state temperature rise due to pn

p1=p

n(11)=9kW Also, A=p

1

T

=9000

50=180 W/oC

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 = 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

+ve

Pm +ve

T -ve

+ve

Pm -ve

T -ve

-ve

Pm +ve

T +ve

-ve

Pm -ve


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4-quadrant operation

m

Te

Te

m

Tem

Te

m

T

Direction of positive (forward)speed is arbitrary chosen

Direction of positive torque will

produce positive (forward) speed

Quadrant 1

Forward motoringQuadrant 2

Forward braking

Quadrant 3

Reverse motoring

Quadrant 4

Reverse braking


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Steady-state stability


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Introduction

DC Motors - 2 pole

Stat

or

Rotor


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Introduction

DC Motors - 2 pole

Mechanical commutator to maintain armature current direction

X

X

X

X

X

Armature mmf produces

flux which distorts main

flux produce by field

Armature reaction


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Introduction

Flux at one side of the pole may saturate

Zero flux region shifted

Flux saturation, effective flux per pole decreases

Large machine employs compensation windings and inter Armature mmf distorts field flux

Armature reaction


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Introduction

Te=kti

a Electric torque

ea=k

E Armature back e.m.f.

Lf Rf

if

vt=R

aia+L

dia

dt+e

a

+

ea

_

LaRa

ia+

Vt

_

+

Vf

_

dt

diLiRv ffff +=


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Introduction

Vt=R

aI

a+E

a

In steady state,

=V

t

kT

RaT

e

(kT)2

Therefore steady state speed is given by,

Vt=Ra ia+L

dia

dt +ea

Armature circuit:


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Three possible methods of

speed control:

Field flux

Armature voltage Vt

Armature resistance Ra


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Electric Drivesnew