Drive Ac Basics 1

8/9/2019 Drive Ac Basics 1

1/24

Dr. Zainal salam; Power

Electronics and Drives

(Version 2),2002, UTMJB

1

Induction Motor Drive Why induction motor (IM)?

Robust; No brushes. No contacts on rotor shaft High Power/Weight, Lower Cost/Power ratios

Easy to manufacture

Almost maintenance-free, except for bearing and otherexternal mechanical parts

Disadvantages

Essentially a fixed-speed machine Speed is determined by the supply frequency

To vary its speed need a variable frequency supply

Motivation for variable-speed AC drives Inverter configuration improved

Fast switching, high power switches

Sophisticated control strategy Microprocessor/DSP implementation

Applications Conveyer line (belt) drives, Roller table, Paper mills,

Traction, Electric vehicles, Elevators, pulleys, Air-conditioning and any industrial process that requiresvariable-speed operation.

The state-of-the-art in IM drives is such that most of theDC drives will be replaced with IM in very near future.


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2

Torque production (1)

Only squirrel-cage IM (SCIM) is considered inthis module

Neglecting all harmonics, the stator establishes aspatially distributed magnetic flux density in theair-gap that rotate at asynchronous speed, 1 :

where

e : supply frequency (in Hz)

p: pole pairs (p=1for 2 pole motor, p=2for 4 pole motor etc)

If the rotor is initially stationary, its conductor issubjected to a sweeping magnetic field, inducingrotor current at synchronous speed.

If the rotor is rotating at synchronous speed (i.e.equals tof1), then the rotor experience no induction.No current is induced in the rotor.

pe =1


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3

Torque production (2)

At any other rotor speed, say wm, the speeddifferential i-2 creates slip. Per-unit slip is

defined as:

pairpole:

frequencyrotor:

frequencysupply::where

; 11

1

p

ps

m

e

em

=

=

Slip frequency is defined as: 2=1-m.

When rotor is rotating at m., rotor current at slipfrequency will be induced.

The interaction between rotor current and air-gapflux produces torque.


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4

Single-phase Equivalent Circuit

(SPEC)

voltage)(phasetageSupply vol:

inductancegMagnetisin:

inductanceleakageRotor:

resistanceRotor:

inductanceleakageStator:

resistanceStator:

1

2

2

1

1

v

L

L

R

L

R

m

R1 L1

V1 Rm Lm VmV2 = nSVm

L2

R2

STATOR SIDE

1:nS


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5

SPEC, referred to stator

From previous diagram, SPEC is a dual frequency

circuit. On the stator is 1 and on the rotor m

Difficult to do calculations.

We can make the circuit a single frequency type, byreferring the quantities to the stator

I1 R1 L1

V1 Rm Lm

L2

R2

S

I2


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6

Rotor current

.alterationresistancerotorhebut with tstator,the toreferrednowarequantitiesthat theNote

phase,rotorin thecurrenttheHence

0)0(

,1wheni.espeed,ssynchronouAt

1

,0wheni.e,standstillAt

:whereEbewillratioturnseffectivesamethewithphaserotorequivalentaninEMFbackthe

thenphase,statorin theEMFbacktheisEIf

22

1

22

1

22

22

12

m

1112

m

12

2

1

jXs

R

EjsXR

sE

jsXR

EI

EE

EEsEE

sEE

+

=

+=

+=

==

=

===

=

=


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7

Performance calculation using SPEC

22

2

22

2

21

12

1

11

11

3:losscopperRotor

3:gap-airtheacrossPower

3:lossCore

3:losscopperStator

currentandvoltagephasebemustand:Note

cos3:PowerInput

RIP

PPPs

RIP

R

VP

RIP

IV

IVP

lrlclsin

g

mlc

ls

in

=

=

=

=

=

=

I1 R1 L1

V1 Rm Lm

L2

R2

S

I2


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8

Performance calculation (2)

ee

ee

e

mm

mm

oe

FWFWosh

g

lrgo

s

RpIT

p

s

RIT

ss

s

sRIPT

PPPP

sPs

sRII

s

RI

PPP

22

2

1

1

22

2

11

1

22

2

22

222

22

2

3

Then,

frequency.supplytheis;But

3

,)1(

Since

)1(3

:torquegnetic)(electromaDeveloped

loss.windageandfriction:;:shaftat thePower

)1()1(3

33

:poweroutputGross

=

=

=

=

=

==

=

=

==

=


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9

Example calculation

A single phase equivalent circuit of a 6-pole SCIM

that operates from a 220V line voltage at 60Hz is

given below. Calculate the stator current, output

power, torque and efficiency at a slip of 2.5%. The

fixed winding and friction losses is 350W. Neglect

the core loss.

I1 R1 X1

V1 Xm

X2

R2

I2

0.2 0.5

20 0.1

0.2

1V 220V line-to line 3

220V127V

3

2.5% 0.025

=

= =

= =


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10

Calculation (solution)

mN

sp

PPT

WPPP

WsPP

W

PPP

WRIP

W

IVP

A

Z

VI

jj

j

jj

jXs

RjXjXRZ

XXX

e

o

m

oe

FWosh

go

lsing

ls

oin

o

oin

o

min

m

.4.78

)025.01)(3/60(2

9611

)1()(

TorqueneticElectromag

%3.8910758

9611

powerInput

powerOutputEfficiency

611,9350961,9

shaftat thePower

961,9)025.01(216,10)1(

powerGross

216,10540758,10

loss)coreg(neglectinrotortosferredPower tran

540)2.0)(30(33

758,10

)20)(cos30)(3220(3cos3

200.30

202.4

3220

202.4

202.0025.0

1.0

2.0025.0

1.0

205.02.0

//)(

20,2.0,5.0

1

21

21

11

11

22

11

21

=

=

==

===

===

===

==

=

===

=

==

=

==

=

++

+

++=

+++=

===


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11

Starting current

For the previous example, Calculate the stating currentwhen motor is first switched on to rated applied

voltage.

motors.inductionstartingtaken whenbeshouldemotors.CarinductionforcommonisThis

current.loadfull

thantimes5aboutiscurrentstartingthat theNote

16776.0

322076.0

202.01.02.01.0205.02.0

1,standstillAt:Solution

11 A

Z

VI

jjjjjZ

s

in

in

===

=

+++++==

=


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12

( )

( )

++

+

==

=

++

+

=

221

21

22

11

2

12

1

222

221

21

22

1

12

m

33

torque,)(developedneticElectromag

phase),(perrotorat thePower

drawnbecanabovecircuitthelarge,isLSince

LLs

RRs

VRPT

s

RIP

LLs

RR

VI

oe

o

Approximate SPEC

+

-

1V ML

1R 1L 2L

s

R2


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13

Single (fixed supply) frequency

characteristics

.0speed,syncat,1standsillat;

:thatNote

below.asshownbecansticscharacteri

slip)(versustorquethe,frequencygiveaFor

1

1

1

==

= sss m

e

m

PLUGGING

TORQUE(+)

MOTORING

emT(max torque or

pull-out torque)

esT (starting torque)

SLIP,s

TORQUE(-)

e

rated slip0 unity slip

(standstill)

GENERATING

e

m

e

m

zero slip

(sync.speed)

SPEED


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14

Single frequency characteristics

CURRENT TORQUE

operating point

(rated torque)

rated slip

rated current

POWERFACTOR

SLIP

1.00

Standstill synchronous speed

EFFICIENCY


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15

Single frequency characteristic

As slip is increased from zero (synchronous), the

torque rapidly reaches the maximum. Then it

decreases to standstill when the slip is unity.

At synchronous speed, torque is almost zero.

At standstill, torque is not too high, but the current

is very high. Thus the VA requirement of the IM is

several times than the full load. Not economic to

operate at this condition.

Only at low slip, the motor current is low andefficiency and power factor are high.


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16

Typical IM Drive System

IDC

Modulation Index,

VDC

n

IM

Supply Rectifier and Filter 3-phase Inverter

CIRCUIT

BLOCK DIAGRAM


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17

Variable speed characteristics

For variable speed operation, the supply is aninverter.

The frequency of the fundamental AC voltage will

determine the speed of IM. To vary the speed of

IM, the inverter fundamental frequency need to be

changed.

The inverter output frequency must be kept close to

the required motor speed. This is necessary as the

IM operates under low slip conditions.

To maintain constant torque, the slip frequency has

to be maintained over the range of supply

frequencies.


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18

Variable voltage, variable

frequency (VVVF) operation

In order for maximum torque production,

motor flux should be maintained at its

rated value.

( )

operation

Volt/HertzconstanttheasknownpopularlyisThis

constant.keptbetohasratiothe

flux,motorhemaintain torder toinTherefore,

44.4or

44.42

1

RMS,In

cos

:isemfbackBut the

sin

11

1

1

1

11111

111

1

fE

Nf

E

NfNE

tNdt

dNe

t

m

mm

m

m

=

==

=

=

=


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19

Constant Torque region

Hence for VVVF operation, there is a need tocontrol the fundamental voltage of the inverter if itsfrequency (and therefore the frequency of the IM)need to be varied.

To vary the fundamental component of the inverter,the MODULATION INDEX can be changed.

The rated supply frequency is normally used as thebase speed

At frequencies below the base speed, the supply

magnitude need to be reduced so as to maintain aconstant Volt/Hertz.

The motor is operated at rated slip at all supplyfrequencies. Hence a constant torque region isobtained.


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20

Constant Torque Region

TORQUE(+)

SPEED

SLIP,s

1f

rated slip

0

2f

3f

4f

5frated torque

54321 fffff >>>>

TORQUE(+)

SPEEDSLIP,s0

rated torque


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21

Constant Power region

Above base speed, the stator voltage

reaches the rated value and the motor

enters a constant power region.

In this region, the air-gap flux decreases.

This is due to increase in frequency

frequency while maintaining fixed

voltage.

However, the stator current is maintained

constant by increasing the slip. This is

equivalent to field weakening mode of a

separately excited DC motor.


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22

Constant Power region

TORQUE(+)

SPEEDSLIP,s Base speed

0

rated torque

TORQUE(+)

SPEEDSLIP,s0

CONSTANT TORQUE

REGION

Base speed

"FIELD WEAKENING"


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23

VVVF Summary

SPEED0

CONSTANT TORQUE

Base speed

CONSTANT POWER

Electromagnetic torque, Te

Terminal (supply)

voltage, V1

slip,s

slip frequency,fs


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(V i 2) 2002 UTMJB

24

Examples

A three-phase 4-pole, 10 horsepower, 460Vrms/60Hz (line-to line) runs at full-loadspeed of 1746 rpm. The motor is fed froman inverter. The flux is made to be constsnt.Plot the torque-speed graphs for the

following frequency: 60Hz, 45 Hz, 30Hz,15Hz.

A three-phase induction motor is using athree-phase VSI for VVVF operation. TheIM has the following rated parameters:

voltage: 415V (RMS)

frequency: 50Hz

slip (p.u) 5%

pole pair 2

If the inverter gives 415V (RMS) with

modulation index of 0.8, calculate the requiredmodulation index if the motor need to beoperated at rotor mechanical speed of 10Hz.

Documents

Drive Ac Basics 1