chapter 3_induction motor cont

8/8/2019 Chapter 3_induction Motor Cont

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CHAPTER 3:INDUCTION MOTOR

(CONT.)

NJFKEE, UMP


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3.5 PRINCIPLE OF OPERATIONThe AC IM is a rotating electric machinedesigned to operate from a three-phase sourceof alternating voltage.

The stator is a classic three phase stator withthe winding displaced by 120.


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Rotating Magnetic FieldWhen a 3 phase stator winding is connectedconnected to a 3 phase voltage supply, 3 phasecurrent will flow in the windingsflow in the windings, which also will inducedinduced 3 phase flux in the stator.These flux will rotate at a speed called a Synchronous Speed, nSynchronous Speed, n ss . The flux is called

Rotating magnetic FieldSynchronous speed = speed of rotating flux:

Where; p = is the number of poles, andf = the frequency of supply

p f n s 120=

3.5 PRINCIPLE OF OPERATION


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At synchronous speed, there is no difference betweenrotor speed and rotating field speed

so no voltage is induced in the rotor bars and no currentflow, hence no torque is developed.

Therefore, when running, the rotor must rotate slower than the magnetic field.

The rotor speed is just slow enough to cause the proper amount of rotor current to flow, so that the resultingtorque is sufficient to overcome windage and frictionlosses, and drive the load.



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Slip,Slip

, ssThis speed difference between the rotor and magnetic field iscalled slip:

Where; n s = synchronous speed (rpm)n r = mechanical speed of rotor (rpm)

under normal operating conditions, s= 0.01 ~ 0.05, which is verysmall and the actual speed is very close to synchronous speed.Note that : s is not negligibles is not negligible

)1( snnORn

nn s sr s

r s =

=



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Rotor SpeedRotor Sp

eed When the rotor move at rotor speed, n r (rps), the stator fluxwill circulate the rotor conductor at a speed of (n s-n r ) per

second. Hence, the frequency of the rotor is written as:

sf

pnn f r sr

=

= )(

f s f iii

ii pnn

f

nn Rotor At

i pn

f

n stator At

Note

r

r sr

p f

r s

s

p f

s

.:)()(

).....(120

)(

:

).....(

120

:

:

120

120

=

=

=

=

=



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Torque producing mechanismTorq ue producing mechanismWhen a 3 phase stator winding is connectedconnected to a3 phase voltage supply, 3 phase current will flowflowin the windingsin the windings, hence the stator is energized.

A rotating flux, is produced in the air gap. The induces a voltage E a in the rotor winding (like atransformer).The induced voltage produces rotor current, if

rotor circuit is closed.The rotor current interacts with the flux ,producing torque. The rotor rotates in thedirection of the rotating flux.



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Q: How to change the direction of rotation?


A: Change the phase sequence of the power supply.


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Note: Never use three-phase equivalent circuit. Always

use per- phase equivalent circuit.

The equivalent circuit always bases on the Y always bases on the Y connection regardless of the actual connection of connection regardless of the actual connection of the motor the motor.

IM equivalent circuit is very similar to the single-

phase equivalent circuit of transformer. It iscomposed of stator circuit and rotor circuit

3.6 EQUIVALENT CIRCUIT


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Step2 Rotor winding is shorted(Under normal operating conditions, the rotor winding is shorted. The slip is s )

Note:the frequency of E 2 is f r =sf because rotor is rotatingrotor is rotating .

f f r



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Step3: Eliminate f 2

Keep the rotor current same:



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Step 4: Referred to the stator side

Note: X 2 and R 2 will be given or measured.

Always refer the rotor side parameters to stator side.R

c represents core loss (of stator side).



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IEEE recommended equivalent circuit

Note:R

c is omitted. The core loss is lumped with therotational loss.



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IEEE recommended equivalent circuit

Note: can be separated into 2 PARTSNote: can be separated into 2 PARTS

Purpose :Purpose :to obtain the developed mechanicalto obtain the developed mechanical

I1 1 R1 X

m X

'2 X

'2 R

s s R 1'21V

s

R2

s s R

R s

R )1(22

2 +=



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Analysis of Induction MachinesFor simplicity, let assume

II ss=I=I 11 , I, I R R =I=I 22

(s=stator, R=rotor)

[ ] Rm sTotal

s s s

cmm

cmcm

R R

R

Z Z Z Z

jX R Z

neglected R jX Z neglected R jX R Z

jX s

R Z

//

;

;;//

;''

+=+=

===

+=

ZZRRZZmm

ZZss

VVs1s1

IIs1s1 IIm1m1 IIR1R1

T

s

s Z

V I

1

1=



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=

=

=

m

RM m

R

RM R

sT

m R RM

Z

V I

Z

V I Hence

V Z

Z Z V

Rules Dividing Voltage

11

11

11

,

//

,

ZZRRZZmm

ZZss

VVs1s1

IIs1s1 IIm1m1 IIR1R1

11

11

,

s Rm

m R

s Rm

Rm

I Z Z

Z I

I Z Z Z

I

Rules Dividing Current

+=

+=

OR

Note : 1hp =746WattNote : 1hp =746Watt

Analysis of Induction Machines



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Example 11. A 4 poles, 3 Induction Motor operates from a

supply which frequency is 50Hz. Calculate:a. The speed at which the magnetic field is rotatingb. The speed of the rotor when slip is 0.04c. The frequency of the rotor when slip is 3%.d. The frequency of the rotor at standstill


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2. A 500hp, 3 6 poles, 50Hz Induction Motor has a speed of 950rpm on full load. Calculate the slip.

Example 2


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3. If the emf in the rotor of an 8 poles Induction Motor has afrequency of 1.5Hz and the supply frequency is 50Hz.Calculate the slip and the speed of the motor.

Example 3


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4. A 440V, 50Hz, 6 poles, Y connected induction motor israted at 135hp. The equivalent circuit parameters are :

R s=0.084 R R=0.066 Xs=0.2 XR=0.165 s = 5% X m=6.9

Determine the stator current, magnetism current and rotor current.

SolutionGiven V=440V, p=6, f=50Hz, 135hp

]63.9685.32,76.1344.170,34.246.177[ A A Aooo

Example 4


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Example 4 (Cont 1 st

Method)


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Example 4 (Cont 2 nd

Method)


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Power Flow Diagram

PP inin (Motor)(Motor)

PP inin (Stator)(Stator)

PP corelosscoreloss(P(P cc ))

PP air Gapair Gap (P(P agag ))

PP

developeddevelopedPP mechanicalmechanicalPP convertedconverted

(P(P mm ))

PP out,out, PP oo

PP stator copper stator copper loss,loss, (P(P scuscu ))

PP rotor copper rotor copper lossloss (P(P rcurcu ))

PP windage, friction, etcwindage, friction, etc (P(P --

Given)Given)

cos3 s s I V

s R

I R R'

'3 2

2

3

c

RM

RV ''3

2 R R R I

s s

R I R R1

''3 2

W hp 7461 =

s s R I 23

PP inin (Rotor)(Rotor)



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Ratio:Ratio:

Pag P rcu Pm

s R

I R

R

''3

2

''32

R R R I

s s

R I R R1

''32

s

11

1 s

1

1 s s1

Ratio makes the analysis simpler to find the value of theparticular power if we have another particular power. For example:

s

s

P

P

m

rcu

=

1

Power Flow Diagram



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Efficiency

Watt xW hp x P

I V P otherwise

P P P

P P P givenare P if

P P

out

s sin

mo

lossesino

losses

in

out

746746

cos3,

,

%100

===

==

=



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Example 5 (Cont from Ex

4)Calculatei. Stator Copper Lossii. Air Gap Power

iii. Power converted from electrical to mechanical power iv. Output power v. Motor efficiency


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Torque-Equation

Torque , can be derived from power equation in term of mechanical power or electrical power .

n P

T Hence

srad nwhereT P Power

260

,

)/(60

2,,

=

==

r

oo

r

mm

n P

T TorqueOutput

n P

T TorqueMechanical

Thus

2

60,

2

60,

,

=

=



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Note that, Mechanical torque can be written in terms of circuitNote that, Mechanical torque can be written in terms of circuitparameters. This is determined by usingparameters. This is determined by using approximation methodapproximation method

...

...

...

)1('

'3

)1('

'3

2

2

==

==

r

R R

r

mm

mr m R

Rm

s s

R I P

T

T P and s s

R I P

+=

22

2

)'()'(

'

2

)(3

R R

R

s

RM m

sX R

sR

n

V T

Torque-speed characteristic

Tm

n s

ss maxmax is the slip for Tis the slip for T maxmax to occur to occur

s=1

Tst

Tmax

s max



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+++

=

=

22

2

)'()'('

602

)(3

1,

R s R s

R

s

s st X X R R

Rn

V T

sTorqueStarting

+++

=

+=

22

2

max

22max

)'()(

1

6022

)(3

)'()(

'

R s s s s

s

R s

R

X X R Rn

V T

X R

R s



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Calculatev. Mechanical torquevi. Output torquevii. Starting torque

viii. Maximum torque and maximum slip

Example 6 (Cont from Ex

4)


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There are 3 types of speed control of 3 phase inductionmachines

i.i. Varying rotor resistanceVarying rotor resistanceii.ii. Varying supply voltageVarying supply voltageiii.iii. Varying supply voltage and supply frequencyVarying supply voltage and supply frequency

3.7 SPEED CONTROL


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Varying rotor resistance

For wound rotor onlySpeed is decreasing

Constant maximum torqueThe speed at which maxtorque occurs changesDisadvantages:

large speed regulationPower loss in R ext reduce the efficiency

T

n s ~n NL

T

n r1n r2n r3 n

n r1 < n r2 < n r3R 1R 2R 3

R 1< R 2< R 3

3.7 SPEED CONTROL


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Varying supply voltage

Maximum torque changesThe speed which at max

torque occurs is constant (atmax torque, X R=R R/s

Relatively simple method uses power electronics

circuit for voltage controller Suitable for fan type loadDisadvantages :

Large speed regulation

since ~ n s

T

n s ~n NL

T

n r1n r2n r3 n

n r1 > n r2 > n r3

V1

V2

V3

V1> V2 > V3

Vdecreasing

3.7 SPEED CONTROL


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The best method since supplyvoltage and supply frequency isvaried to keep V V / / f f constant

Maintain speed regulationuses power electronics circuitfor frequency and voltagecontroller Constant maximum torque

Varying supply voltage and supply frequency

T

n NL1

T

n r1n r2n r3 n

f decreasing

n NL2n NL3

3.7 SPEED CONTROL


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THANK YOU FOR YOUR ATTENTION!!

HAVE A NICEHOLIDAY!

chapter 3_induction motor cont

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