Induction Motors-scalar Control

INDUCTION MOTORS-SCALAR CONTROL

Considering two pole machine with three phase windings on the stator

ROTATING MAGNETIC FIELD-ANALYTICAL METHOD

Considering two-pole machine with three phase windings on the stator.

R lt t MMF t i t i th i d fi d b l θResultant MMF wave at any point in the air gap, defined by an angle θm.The origin of the angle θm can be chosen to be the axis of phase u. At anyinstant of time, all three phases contribute to the air gap mmf along the pathdefined by θ The mmf along θ is

1

defined by θm. The mmf along θm is)()()()( mwmvmum FFFF

INDUCTION MOTORS-SCALAR CONTROLROTATING MAGNETIC FIELD-ANALYTICAL METHOD

2


Where N is the effective number of turns in phase U and i is current in phase U


Where N is the effective number of turns in phase U, and iu is current in phase U

Because the phase axes are shifted from each other by 120 electrical degrees

)cos()( mumu iNF

Because the phase axes are shifted from each other by 120 electrical degrees,the contributions from phases V and W are, respectively

)º120cos()( iNF )º120cos()()120cos()(

mwmw

mvmv

iNFiNF

The resultant MMF at point θ is:p

)120cos()120cos()cos(),( ºº mwmvmum tiNtiNtiNtF

Th t i i iThe currents iu, iv, iw, are:

º120cos

cos

tItitIti smu

3

º120cos

120cos tIti

tIti

smw

smv


)120cos()º120cos()120cos()º120cos()cos()cos(),( ºº msmmsmmsmm tINtINtINtF


BABABA cos21cos

21coscos

3

In general for a P-pole machine that is 2p poles we have:

)cos(23),( msmm tINtF

In general for a P-pole machine, that is 2p poles, we have:

)cos(23),( msmm ptI

pNtF

gtFtB

gtFtH m

mm

m),(,),(, 0

Resultant mmf wave in the air-gap rotating at the constant angular velocity Ωs

INtFpt 3)(0

pdtdt

p

Ip

tFpt

smsm

mmms 2),(0

4

mprnpf

pf

srad

p

pp

ssss

s ..60302



5


The speed of the magnetic rotating field is the synchronous speed For a

SYNCHRONOUS SPEED, ASYNCHRONOUS SPEED, AND SLIP SPEED

The speed of the magnetic rotating field is the synchronous speed. For a induction motor with P poles, the synchronous speed is given in r/min as:

fffrev sss 6060120

where, fs is the stator frequency in Hertz, and ns is the synchronous speed in r/min

pf

Pf

Pfn sss

s

2min

r/min.

However, the rotor rotates at a speed slightly slower than the synchronous speed. This difference between speeds is called the slip speed and it is given as:

h i th h i l d i / i f th t d i th li d

msslip nnn

where, nm is the mechanical speed in r/min of the rotor, and ns is the slip speed in r/min. Moreover, the slip speed can also be defined in a per unit system as the slip, s, as given in:

6

s

ms

nnns

INDUCTION MOTORS-SCALAR CONTROLSYNCHRONOUS SPEED, ASYNCHRONOUS SPEED, AND SLIP SPEED

7



It also can be expressed as

pnnsm

s

msms

It also can be expressed as

m

d

revnmin

pn

ssss

So:ss

m

fs

rad

2

Frequency of the rotor voltages and currents

smsm snsn 11

Frequency of the rotor voltages and currents

mss

sms

s

ffPPPs

8

srsssmsr fsfsPssPP

222


We have seen that the speed of the magnetic rotating field produced by a


We have seen that the speed of the magnetic rotating field produced by a currents of fs, respect to the reference system of the stator (fixed) is:

fn ss

60

ps

On the other hand, the currents of the rotor of frequency sfs produces another rotating magnetic field, which respect to the reference system attached to the g g p yrotor turns at:

ss

sr nsp

fsn 60

With the rotor revolving in the same direction of rotation as the stator field, the rotor currents produce a rotating flux wave rotating at sns with respect to the rotor in the forward direction.

With respect to the stator, the speed of the flux wave produced by the rotor currents (with frequency sfs ) equals:

9

sssms nsnnsnns 1

INDUCTION MOTORS-SCALAR CONTROLTORQUE IN ROTOR

The N-S poles created by the stator sweeps to the right at synchronous speed n The flux per pole is distributed sinusoidally The rotor is also moving to thenS. The flux per pole is distributed sinusoidally. The rotor is also moving to the right at a speed nm

A voltage E=B·l·(s-m ) is induced in each conductor while it is being cut by the mutual flux that crosses the air gap.

The induced voltage immediately produces a current i, which flows down the conductor undeaneath the pole-face through the end bars and back through theconductor undeaneath the pole face, through the end bars, and back through the other conductors.

Because the current-carrying conductor lies in the magnetic field of the t t it i h i l f F l i Bpermanent magnet, it experiences a mechanical force F=l·i·B;

the force always acts in a direction to drag the conductor along with the magnetic field.g

10


TORQUE IN ROTOR

The inducted voltage E its in phase with the flux, but the current I lags an angle due to the RL load due to the bars. This lagging reduces the torque for the samecurrent.

11


TORQUE IN ROTOR

Because the stator and rotor fields each rotate synchronously they are

12

Because the stator and rotor fields each rotate synchronously, they are stationary with respect to each other and produce a steady torque, thus maintaining rotation of the rotor.


Three-phase windings shifted 120eº

TORQUE IN ROTOR

p g e

Rotating field at 60·fs/p

Induces voltage in the rotor bars

A current flows through the rotor bars

Force applied on the rotor bars BlIF

The rotor produces mechanical torque

13The IM rotates


The rotor currents, Ir produce additional air gap flux, which is 90º out of phase of the magnetizing flux But the stator voltage is applied externally and it is

INDUCTION MOTOR EQUIVALENT CIRCUIT

the magnetizing flux. But the stator voltage, is applied externally and it is proportional to and 90º out of phase of the air gap flux. Additional currents, will flow in the stator windings in order to cancel the flux due to the rotor currents.

Per phase transformer-like equivalent circuitPer phase transformer-like equivalent circuit.

14Purely resistive conductors


INDUCTION MOTOR EQUIVALENT CIRCUIT

Effect of Rotor Inductance on Torque:C t l i k t ftCurrent lags since peak current occurs afterstator pole passes by reducing the torque

Iron losses are no taken into account,i.e. Ife=0

15

INDUCTION MOTORS-SCALAR CONTROLPER PHASE INDUCTION MOTOR EQUIVALENT CIRCUIT

16


TORQUE

2'1'3 IsRP

222''3'1'3'1'3

3

rr

rrrrmi

rrmi

Is

RIs

sRIs

sRPT

Is

RP

21

''

1f

r

ssmm

mii

R

UI

ss

T

22

22

'3''3

''

frr

rsr

s

URIR

XXs

RR

2

2

1

''

33

rsr

sss

r

i

XXs

RR

Us

IsT

22

21

'

'3 fr

iR

Us

R

T

17

2'rs

rss XX

sRR


160

T [N ]

Rs=0.2268; Rr’=0.12528; Xs=0.51252; Xr’=0.76878; Rfe=185.1; Xm=9.8554; U1L=230

140

Torque [Nm]Output power [CV]Current [A]Efficiency [%]

100

120y [ ]

Power Factor*100

60

80

40

60

0

20

18

0 100 200 300 400 500 600 700 800 900 10000

Speed, rpm


160

Torque [Nm]

Rs=0.2268; Rr’=0.12528; Xs=0.51252; Xr’=0.76878; Rfe=185.1; Xm=9.8554; U1L=230

140

Torque [Nm]Output power [CV]Current [A]Efficiency [%]

100

120 Power Factor *100

80

40

60

20

19

900 920 940 960 980 10000Speed, rpm


150Speed-torque curve

100

50

m]

0

Torq

ue [

N

-50

T

150

-100

20

-1500 -1000 -500 0 500 1000 1500 2000 2500 3000-150Shaft speed [rpm]


In the brake region, the rotor rotates in the opposite direction to that of the airgap flux so that s>1. This condition may arise if the stator supply phasesequence is reversed when the rotor is moving.In the regenerating region the motor as acts as a generator . The rotor runs

> i th di ti th t t fi ld <0 Th ti linm>ns ,in the same direction as the stator field, so s<0. The negative slipcorresponds to negative Rr/s . The positive resistance consumes energy duringmotoring, but the negative Rr/s generates energy and supply it to the source.

d it i

21

0 ≤ nm < nss (0,1)

nm > nss < 0

nm and ns opposite signs > 1


r IRIsRIsR 222''3'1'3'1'3

SPEED CONTROL-VARIABLE FREQUENCY DRIVES

rsr

s

f

rs

r

s

rr

m

rr

m

mii

XXs

RR

UII

ss

Is

RIs

RPT

22

1

'''

3

1

33

frfr Us

RsU

sR

T

212

21

'3'3

rsr

ssrsr

ss

i

XXs

RR

sss

XXs

RR

sT2

222

2

''''

rs

f

s

f

rs

f

rsrss

fr

i sRf

UpsR

UsR

UXXsRRs

URsT2

12

12

1222

21

'23

'

3'

3''

'3

For low slip, i.e., very close to rated

speed

fff

rsr

rsrsrss

UpfUpUp

fRp

222 333 fU2

3

speed.

22

r

s

f

rs

r

sr

f

sr

f

i ff

URp

ff

fRUps

fRUpT 111

'23

'23

'23

rs

f

ri f

fU

RpT 1

'23

INDUCTION MOTORS-SCALAR CONTROLDIFFERENT SCENARIOS OF CONSTANT TORQUE MODE

1 kW, 4 pole, 1740 rev/min

416 V, 3-phase, 60 Hz

RATED OPERATING MODE

The frequency applied to the stator is 60 Hz, so synchronous speed is 1800rev/min. The motor develops rated torque at 1740 rev/min, which corresponds toa slip of (1800 1740)=60 rev/min

23

a slip of (1800-1740)=60 rev/min.

The rotor frequency is fr=sf=((1800-1740)/1800)*60 Hz =2 Hz


OPERATION AT 6.1 HZ

The frequency applied to the stator is 6.1 Hz, so synchronous speed is 183rev/min. In order to produce the same torque as before the current in rotorbars must be exactly as it was before in magnitude, frequency and phase.bars must be exactly as it was before in magnitude, frequency and phase.This is realized when the slip speed is 60 rev/min. Thus full-load torque isproduced when the rotor turns at 183-60= 123 rev/min.

The ind ced oltage in the stator is no less than before beca se the fl is onl

24

The induced voltage in the stator is now less than before because the flux is onlyturning at 183 rev/min. The value is 240V/1800 rev/min*183 rev/min= 24.4 V ;frequency 6.1 Hz


MOTOR STOPPED

The frequency applied to the stator is 2 Hz, so synchronous speed is 60 rev/min. In order to produce the same torque the slip speed must be again 60 rev/min This means that n=0

25

rev/min. This means that n 0.

The induced voltage in the stator is 240V/1800 rev/min*60 rev/min= 8 V ; frequency 2 Hz


MOTOR OPERATING AS A BRAKE

The frequency applied to the stator is 0.5 Hz, so synchronous speed is 15 rev/min. In order to produce the same torque the slip speed must be again 60 rev/min This means (15-60)=-45 rev/min The flux and rotor turn in opposite

26

rev/min. This means (15 60) 45 rev/min. The flux and rotor turn in opposite directions.

The induced voltage in the stator is 2V ; frequency 0.5 Hz


ROTOR EXCITED BY DC

The frequency applied to the stator is 0 Hz, so synchronous speed is 15 rev/min. Nevertheless, the rated torque can again be obtained provided the relative speed is 60 rev/min This happens when the rotor turns at 60 rev/min in

27

relative speed is 60 rev/min. This happens when the rotor turns at 60 rev/min in either direction. Again the rotor acts as a brake.

The induced voltage in the stator is 0V ; frequency 0Hz


OPERATION ABOVE BASE SPEEDOPERATION ABOVE BASE SPEED

The frequency applied to the stator is 150 Hz, so synchronous speed is 4500 rev/min. In order to produce the same torque the slip speed must be again 60

/ i Th t i t i t 4440 / i

28

rev/min. The rotor is turning at 4440 rev/min.

The induced voltage in the stator is 240/1800*4500= 600V ; frequency 150 Hz


400Induction motor torque-speed for variable frequency

Frequency increment: 4.6667 HzBase curve(solidline): 50 Hz

300

350Base curve(solidline): 50 Hz

250

N-m

150

200

Torq

ue, N

100

0 500 1000 1500 2000 25000

50

29

Speed, rpm


305.5 kW, 50 Hz, 1440 rev/min4-poles


31


32

INDUCTION MOTORS-SCALAR CONTROLSPEED CONTROL-VARIABLE FREQUENCY DRIVES

F i i t b d h t i t t d b b d

33

From minimun to base speed where torque remains constant, and beyong basedspeed where due to a constant power limitation torque decreases with frequencyincrease.


THE REQUIREMENTS FOR STOPPING

DYNAMIC BRAKINGDYNAMIC BRAKING

The next figure shows the change in the motor torque when the converter output frequency issuddenly reduced from f0 to f1. The slip changes from being positive (motoring) to negative(generating) and the direction of energy flow is reversed kinetic energy is converted to electrical(generating) and the direction of energy flow is reversed, kinetic energy is converted to electricalenergy in the motor, which is then transferred from the motor to the converter.

34

Torque–speed characteristics of an induction motor when frequency is reduced from f0 to f1




On PWM converters, which use a diode rectifier bridge, the braking current is blocked fromreturning to the mains power supply. Therefore, unless some mechanism is provided to absorb thisenergy during braking the voltage on the DC bus will rise to destructive levelsenergy during braking, the voltage on the DC bus will rise to destructive levels.With dynamic braking, the braking energy is dissipated in a braking resistor connectedacross the DC bus of the converter. As described above, braking is achieved by reducing theinverter output frequency to be less than the actual rotor speed. The slip can be optimized to giveas high a torque per ampere as for motoringas high a torque per ampere as for motoring.Power flow is from the motor back through the inverter to the DC bus. The braking energy cannotbe returned to the mains supply because the input rectifier can only transfer power in onedirection. Instead, the energy is absorbed by the DC capacitor, whose voltage rises. To prevent theDC bus voltage rising to a dangerously high level the capacitor needs to be periodicallyDC bus voltage rising to a dangerously high level, the capacitor needs to be periodicallydischarged. This is done by means of a dynamic brake module shown in the next figure, consistingof a power electronic switch, usually an IGBT or BJT, and a discharge resistor connected acrossthe DC link capacitor.

35




36

PWM AC converter with a DC link dynamic brake




The IGBT or BJT is controlled by a hysteresis circuit to turn on when the capacitor voltage is too highand turn off when the voltage drops below a certain level.Alternatively the IGBT may be switched on and off at constant frequencyAlternatively, the IGBT may be switched on and off at constant frequency,with duty cycle varying linearly between 0% and 100% as bus voltage changes overspecific range.

37

DC bus voltage with hysteresis control on a dynamic brake




The switching level of the braking IGBT should be chosen to be higher than the mains supply when itis operating at highest voltage tolerance, but below the maximum safe switching voltage of theinverter components. In practice, for a converter connected to a 3-phase 415 volt supply, with ainverter components. In practice, for a converter connected to a 3 phase 415 volt supply, with anominal DC peak voltage of 586 V, the switching level would have to be set at least 10% above thisat 650 V, but below 800 V, which is the maximum safe operating voltage of the DC bus. A practicalswitch-on level is typically 750 V, with the hysteresis between the upper and lower level being 20 Vto 30 V lower.

A 22 kW VSD and motor combination must provide 100% rated motor torque while braking. The maximum braking duty is 3 seconds in every 10 seconds.Assume that the bus voltage during braking is 650 V DC and the DC bus over-voltage trip level is set

o 30 o e

Assume that the bus voltage during braking is 650 V DC and the DC bus over voltage trip level is set at 700 V DC. What braking resistor should be used for the application?

To achieve rated torque while braking, the resistor must absorb a full 22 kW when the motor is at full speed. Therefore, the maximum DC bus current will be roughlyspeed e e o e, e a u C bus cu e be oug y

38




To absorb 34 amps when the bus voltage is at 650 V DC, the braking resistor will need to have aresistance of 650/34 = 19 ohms. If braking only occurs for 3 sec in every 10 sec, then the duty willbe 30%. The power rating of the braking resistor will then be 30% of 22 kW, which is approximatelybe 30%. The power rating of the braking resistor will then be 30% of 22 kW, which is approximately7 kW continuous. Care must be taken to allow adequate excess power rating when selecting abraking resistor, as the instantaneous power is very high and hot spots can cause prematurefailure.The maximum transistor current will occur at the maximum DC bus voltagee a u a s s o cu e occu a e a u C bus o age

Allowing a safety margin, a braking transistor rated at 50 amps would be selected.

39

INDUCTION MOTORS-SCALAR CONTROLExamples, Three-Phase non filtered diagram

This figure shows angexample of 3P nonfiltered PWM circuitcreated to control aninduction motor Theinduction motor. Theconfiguration shown hastriplicated the opamcircuit explained before,pthere is one modulatorper phase but acommom carrier.

Note that every phase uses two transistors (IGBT) per phase to generate the final signal not four as show before called half bridge configuration

40

signal, not four as show before, called half-bridge configuration.

INDUCTION MOTORS-SCALAR CONTROLExamples, Three-Phase non filtered diagram

Output, Sinusoidal waveform current (lik 3P AC

Input, DC line. Voltage

(like a 3P AC current source)

Voltage does not

varies though

titime.

Input,Three

Output, motor torque. Depends

sinusoidal signal waveform

q pon all Inputs.

Input, One triangular signal waveform generatorC) Frequency variation Higher PWM signal resolution Output signal with less noise, higher frequency noise.

A) Frequency variationOutput frequency variation

B) Amplitude variation

41

Smaller torque variations.B) Amplitude variation

Output amplitude variation

INDUCTION MOTORS-SCALAR CONTROLA) Examples, Frequency Variation

Resultant currents

Voltage signals

(Intensity)

g g

Variation of Modulator signals in PWM generator

60 Hz signal 0,2sin PWM generator.

42

90 Hz signal 0,2s 30 Hz signal 0,2s

INDUCTION MOTORS-SCALAR CONTROLB) Examples, Amplitude Variation

Resultant currents

Voltage signals

(Intensity)

(A*=0,9 V)

(Voltage)Variation of Modulator signals in PWM generator

Original Voltage Signal 0,2sin PWM generator.

*RMS values

43

Half Amplitude (A*=0,45 V) 0,2s Double Amplitude (A*=1,8 V) 0,2s

INDUCTION MOTORS-SCALAR CONTROLC) Examples, Carrier frequency variation

Variation of Carrier signals in PWM generator

! Transistors must support higher

it hi fPWM generator.

Double carrier frequency

Resultant currents (A) Resultant currents (A)

switching frequency

( ) Resultant currents (A)

Carrier and A-Phase signal (V) Carrier and A-Phase signal (V)

1,5KHz 3KHz

Source signals

(Frequency)

A-Phase PWM signal (V) A-Phase PWM signal (V)

44

INDUCTION MOTORS-SCALAR CONTROLC) Examples, Carrier frequency variation

Variation of Carrier signals in PWM generatorPWM generator.

Half carrier frequency

Resultant currents (A) Resultant currents (A)( ) Resultant currents (A)

Carrier and A-Phase signal (V) Carrier and A-Phase signal (V)

1,5KHz 750Hz

A-Phase PWM signal (V) A-Phase PWM signal (V)

45

INDUCTION MOTORS-SCALAR CONTROLC) Transistor Switching

VV

V VVV V

A,B,C A,nB,nC NA,nB,nC NA,B,C NA,nB,C

A B CA No B No CNo A No B No CNo A B CNo A No B C

V V

V V VV V

46

A B C A No B No C No A No B No C No A B C No A No B C


SPEED REGULATION- FREQUENCY-VOLTAGE CONTROL

This figure depicts three tension (3 phases) with sThis figure depicts three tension (3 phases) with sfrequency and Vs with a minimum value for near-zerofrequencies. Inputs are come from speed and it not requiresfeedback.

47

Open loop volts/Hz control with voltage-fed inverter


SPEED REGULATION- FREQUENCY-VOLTAGE CONTROL

Cl d l t l ith lt /H t l d li l ti

48

Closed loop control with volts/Hz control and slip regulation


49


50


51


52


53


Example.Load with constant torque.q

P= 15 kW->n= 2600 rev/min

Minimum speed= 867 rev/minMaximum speed= 2600 rev/min

NmT 55

6022600

15000

60

54


1. Case

2-poles motor ∆/Y 230/400The maximum voltage supply is 380/400 VConnected in Y, constant flux at 230V/50 Hz

The maximum and minimum supply frequency is:

pfn 60

Hzf

12600

45.1460

1867min

Hzf 3.4360

12600max

55


fto min

fto

NmNmNmTn

uM

M 51.7078.0155155

max

NmNmNmTn

uM

M 3.5796.0155155

Si [ ]Number

kWNmP Hzu 14.22602300051.70)50(

AM180 MO 2

Size [mm] poles

56

Hzu 60)50(


2. Case

4-poles motor ∆/Y 230/400The maximum voltage supply is 380/400 VConnected in Y, constant flux at 230V/50 Hz


f60

Hzf

pfn

9.282867

60

i

Hzf

Hzf

67.8660

22600

9.2860

max

min

57


fto min

fto

NmNmNmTn

uM

M

11

7.6485.0155155

max

NmNmNmTn

uM

M 24.11249.0155155

Number

2

Size [mm] poles

58

kWNmP Hzu 62.17602150024.112)50( AM180 MO 4


3. Case

4-poles motor ∆/Y 230/400The maximum voltage supply is 380/400 VConnected in ∆, constant flux at 400V/87 Hz


f60

Hzf

pfn

9282867

60

Hzf

Hzf

67.8660

22600

9.2860

max

min

60

59


fto min

fto

NmNmNmTn

uM

M

11

7.6485.0155155

max

NmNmNmTn

uM

M 7.6485.0155155

Number

2

Size [mm] poles

60

kWNmP Hzu 16.1060215007.64)50( AM160 MO 4

Induction Motors-scalar Control

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