CHAPTER 3 INDUCTION MOTOR.pdf

8/12/2019 CHAPTER 3 INDUCTION MOTOR.pdf

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CHAPTER 3

IDUCTIO MOTOR

3.1 ITRODUCTIO TO IDUCTIO MOTOR

3.2 COSTRUCTIO

3.2.1 TYPES OF ROTOR

a. SQUIRREL-CAGE ROTOR

b. WOUND ROTOR

3.2.2 STATOR

3.3 PRODUCTIO OF ROTATIG MAGETIC FIELD

3.4 PRICIPLE OF OPERATIO

3.5 SLIP AD SYCHROOUS SPEED

3.6 PER PHASE EQUIVALET CIRCUIT

3.7 POWER FLOW DIAGRAM

3.8 IDUCTIO MOTOR TESTS

3.8.1 NO LOAD TEST/ RUNNING LIGHT TEST

3.8.2 BLOCKED/LOCKED ROTOR TEST

3.8.3 DC RESISTANCE TEST

3.9 EFFICIECY

3.10 TORQUE EQUATIO

3.10.1 MECHANICAL TORQUE

3.10.2 MAXIMUM TORQUE

3.10.3 STARTING TORQUE

3.11 STARTIG METHOD AD SPEED COTROL

3.11.1 USING PRIMARY RESISTORS

3.11.2 USING STAR-DELTA STARTER

3.11.3 USING AUTO TRANSFORMER


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3.1 ITRODUCTIO TO IDUCTIO MOTOR

Definition of Induction:

The process by which an electromotive force is produced in a circuit by varying the

magnetic field linked with the circuit.

Induction motors are the most commonly used electric motors.

Although it is possible to use an induction machine as either a motor or a generator, it has

many disadvantages and low efficiency as a generator and so is rarely used in thatmanner. The performance characteristics as a generator are not satisfactory for most

applications.

For this reason, induction machines are usually referred to as induction motors.

AC current supplied to the stator winding produces a flux through the air gap that induces

currents in the rotor windings.

Rotor receives electric power by induction in exactly the same way as thesecondary of 2 winding transformer.

Can be treated as a rotating transformer, one in which primary winding isstationary (stator) but the secondary is free to rotate (rotor).

Most appliances, such as washing machines and refrigerators, use a single-phaseinduction machine

For industrial applications, the three-phase induction motor is used to drivemachines

Advantages

Very simple and extremely rugged Low cost and very reliable Requires minimum of maintenance


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Disadvantages

Speed cannot be varied without sacrificing some of its efficiency. Speed decreases with increase in load

3.2 COSTRUCTIO

A machine is called induction machines because the rotor voltage (which produces the

rotor current and the rotor magnetic field) is induced in the rotor windings instead of

being physically connected by wires.

The distinguishing feature of an induction machine is that no DC field current is required

to run the machine.

Although it is possible to use an induction machine as either a motor or a generator, it has

many disadvantages as a generator and so is rarely used in that manner. For this reason,

induction machines are usually referred to as induction motor.

Two sets of electromagnets are formed inside any motor. In an AC induction motor, one

set of electromagnets is formed in the stator because of the AC supply connected to the

stator windings. The alternating nature of the supply voltage induces an Electromagnetic

Force (EMF) in the rotor (just like the voltage is induced in the transformer secondary) as

per Lenzs law, thus generating another set of electromagnets; hence the name

induction motor. Interaction between the magnetic field of these electromagnets

generates twisting force, or torque. As a result, the motor rotates in the direction of the

resultant torque.

An induction motor consists of two main parts: stator and rotor. It has the same

physical stator as a synchronous machine but with different rotor construction. There are

two types of induction motor rotors that can be placed inside the stator, i.e. squirrel-cage

rotor and wound rotor.


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3.2.1 TYPES OF ROTOR

a) Squirrel-Cage Rotor

Squirrel-cage rotor, as shown below, consists of a series of conducting bars laid

into slots carved in the face of the rotor and shorted at either end by large shorting

rings.

Fig. 3.0: Example of Squirrel-Cage Rotor

The rotor is cylindrical and is made of conducting bars short circuited at both ends It is also known as brushless induction motor. It is more rugged and since there are no brushes it is safer in combustible

environment.

b) Wound Rotor

Wound rotor, as shown below has a complete set of three-phase windings similar to

stator windings. Usually, it is Y-connected and the rotor coils are tied to the slip rings.


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Fig. 3.1: Wound Rotor

The rotor is cylindrical and is made up of a three phase windings with terminalsbrought out to slip rings

Wound rotor induction motors are also known as a slip-ring motors This type is the more complicated of the two type but it has a higher starting

torque and is more controllable

3.2.2 STATOR

The stator is made up of several thin laminations of aluminum or cast iron. They are

punched and clamped together to form a hollow cylinder (stator core) with slots as shown

in Fig. 3.2. Coils of insulated wires are inserted into these slots. Each grouping of coils,

together with the core it surrounds, forms an electromagnet (a pair of poles) on the

application of AC supply.

The number of poles of an AC induction motor depends on the internal connection of thestator windings. The stator windings are connected directly to the power source.

Internally they are connected in such a way, that on applying AC supply, a rotating

magnetic field is created.


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Fig. 3.2: A Typical Stator

3.3 PRODUCTIO OF ROTATIG MAGETIC FIELD

When a three phase stator winding is connected to a three phase voltage supply, three

phase currents will flow in the windings which induce three-phase flux in the stator. This

flux will rotate at a speed called as synchronous speeds

. The flux is called as rotating

magnetization field. The mathematical equation is given as:

p

f120

s= where =f the supply frequency

=p no. of poles in the

machine/motor

The currents that flows in the stator are spaced 120 each other. Graphical representation

is shown in Fig. 3.3.


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Fig. 3.3: 3-Phase Current

( ) tsinitimR =

( ) ( )= 120tsinitimY

( ) ( )+= 120tsinitimB

3.4 PRICIPLE OF OPERATIO

When a three phase current flow in a three-phase winding, rotating magnetic field (flux)

will be produced. The flux has constant magnitude and is distributed in sinusoidal form.

This flux will induce voltage in the rotor conductor by Flemings Right Hand Rule. By

Faradays Law, if the rotor winding is short-circuited, rotor current will flow in it. The

reaction between rotor current and stator flux causes the rotor to rotate in the same

direction as the stator flux.

An induction motor with 2 poles can be taken to explain this phenomena. Conductor A

will be located under north pole while conductor B will be located under south pole as

illustrated in Fig. 3.4. The flux rotates in the clock-wise direction (towards the right).lf

the flux is taken as the reference, the conductors A and B are likely to move to the left.

Then, from the Right Hand Fleming, voltage or current will be induced as shown in

Fig. 3.5.


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Fig. 3.4: Conductor A is located under north pole and conductor B is located under south pole

Fig. 3.5: Right Hand Fleming Fig. 3.6: Ampere Right Hand Rule

The same process happens to conductor B.

As shown in Fig. 3.6, when current flows in the rotor circuit, flux will be induced and the

direction is anti-clock-wise. This is called Ampere Right Hand Rule. The interaction

between flux produced by the rotor current and the rotating flux will induce torque on the

rotor conductor that acts to the right. This torque causes the rotor to rotate clockwise. The

illustration is shown in Fig. 3.7.


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Fig. 3.7: Interaction between Rotor Current Flux and Rotating Flux

Conclusion:

Rotating field will cause the rotor to rotate at the same direction as the stator flux. The

torque direction is independent upon the conductor position. Torque direction is always

the same as the flux rotation.

At the time of starting the motor, rotor speed = 0. The rotating magnetic field will cause

the rotor to rotate from 0 speeds to a speed that is lower than the synchronous speed. If

the rotor speed is equal to the synchronous speed, there will be no cutting of flux and

rotor current equals zero. Therefore, it is not possible for the rotor to rotate ats

.

3.5 SLIP AD SYCHROOUS SPEED

Slip is defined as the difference between synchronous speed (magnetic fields speed) and

rotor speed:

s

rs

s

= .. (3.0)

where: =s

synchronous speed in rev/min.

=r

rotor speed in rev/min.


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From Eqn. 3.0, the rotor speed can be derived as ( )s1sr = . Slip can also be

represented in percent. When the rotor move atrn rev/sec (rps), the stator flux will

circulate the rotor conductor at a speed of ( )rs

nn per second. Therefore, the frequency

of the rotor emf,r

f is written as:

( )pnnfrsr

=

sf=

where:

=s slip

=f supply frequency.

The rotor therefore runs at a speed slightly less than the synchronous speed the difference

being called slip speed.

Slip speedrs

=

3.6 PER PHASE EQUIVALET CIRCUITThe per-phase equivalent circuit of a three-phase induction motor is just like a single

phase transformer equivalent circuit. The difference is only that, the secondary winding is

short-circuited unlike in the transformer it is open-circuited as a load is to be connected

later. The per-phase equivalent circuit is illustrated in Fig. 3.8 below.

Fig. 3.8: Per-Phase Equivalent of 3-Phase Induction Motor.


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The per-phase equivalent circuit referred to the stator winding is shown in Figure 742.

This equivalent circuit is categorized into two types: [i] actual equivalent circuit and [ii]

approximate equivalent circuit.

Fig. 3.9: Per-Phase Equivalent Circuit Referred to Stator Winding.

From the equivalent circuit;

=1

I Stator phase current. :2o1

III +=

= Stator line current (for stator Y-connection)

=3

IL , where =

LI stator line current (for stator: - connection)

=2

I Rotor current referred to stator winding

=oI No-load current, mco III +=

=c

I Core current

=m

I Magnetizing current

For approximate equivalent circuit;

2o1 III +=

mco III +=

( )

c

phs

cR

EI = ;

( )

m

phs

mjX

EI = ;

( )

( )121

2

phs

2

XXjRs

R

EI

++

+

= ... (3.1)

This model is normally used for analysis purposes for simplicity.


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3.7 POWER FLOW DIAGRAM

Power flow diagram is actually a flow of power right from the input to the output.

Fig. 3.10: Power Flow Diagram

Fig. 3.10 shows the power flow diagram while Fig. 3.11 illustrates the components that

involve in the power losses calculation.

Fig. 3.11: Components Involved in Power Flow Diagram


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From the circuit shown in Fig. 3.11, the power equations can be derived as follows:

[i] SCL occurs ats

R , then ( ) s

2

2 RI3SCL= . (3.2)

[ii] RCL occurs atR2

'RR = , then ( ) 2

2

2 RI3RCL= . (3.3)

[iii]m

P occurs at ( )s1s

R2 , then ( ) ( )s1

s

RI3P 2

2

2m = . (3.4)

[iv] RIPoccurs ons

R2 , then : ( )

s

RI3RIP 2

2

2= . (3.5)

From Eqn. (3.2...3.5), we can derive the power equations in terms of slip and power,

then:

From Eqn. 3.4 : ( ) ( )s1s

RI3P 22

2m = = ( )

s

s1RCL = ( )s1RIP (3.6)

From Eqn. 3.5 : ( )s

RI3RIP 2

2

2= =

( )s1P

m

=

s

RCL (3.7)

From these equations, we do not need to recalculatem

P andRIP, if RCL is known

provided that the value for slip is known.

The input power comes from the stator input, then:

cosIV3cosIV3PphphLLin

==

WhereLV and

LI are line voltage and line current respectively.

phV and

phI are phase voltage and phase current respectively.

is the angle between voltage and current.


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3.8 IDUCTIO MOTOR TESTS

The equivalent circuit of an induction motor is a very useful tool for determining the

motors response to change in load. There are three tests that are normally being carried

out for a 3-phase induction motor, namely:

i) No load test/ running light testii) Blocked/Locked rotor test

iii) DC resistance test [ this is optional: used to determine stator resistance ]

3.8.1 O LOAD TEST/ RUIG LIGHT TESTThe test set-up is shown in figure below. In this test, the motor is running at rated voltage

and no-load is connected to the motor. Readings are taken on the stator part are:

LV : line to line stator voltage

LI : stator line current

LP : 3-phase input power

Fig. 3.12: Test Set-Up for No Load Test

CBAL IIII ++=

From this test, we can determine the value ofCR and mX . This is corresponding to the

measurement as shown in figure below.


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Fig. 3.13: Equivalent Circuit of Induction Motor

Calculation ofC

R and mX

LL

L

L

LLL

L

L

oc

s

IV

Pcos

cosIV3

PP

3

VE

=

==

=

From phasor diagram shown below,

Fig. 3.14: Phasor Diagram from No-Load Test


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LLC

LLm

cosII

sinII

=

=

m

L

m

C

L

C

I

VX

I

VR

=

=

3.8.2 BLOCKED/LOCKED ROTOR TEST

In this test, the rotor is locked (not moving). The set-up is shown on figure below.

Fig. 3.14: Test Set-Up for Blocked Rotor Test

Power, voltage and current readings are taken.

BRP : Blocked rotor power

BRI : Line current

BRV : Line voltage

From this test, we can determine the value of RBRand XBR.


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rSBR

2

BR

2

BRBR

BR

BR

BR

2

BR

BR

BR

BR

2

BR

BR

BR

'RRR

RZX

I

VZ

IPR

RI3

PP

+=

=

=

=

==

The value rR' of can be calculated if sR is known from the DC test.

3.8.3 DC RESISTACE TESTThis test is performed to determine stators resistance sR . There are two possible

connections: Y or connection.

Fig. 3.15: DC Resistance Test (Y-Connection)

dcs

dcs

dc

dc

RR

RRIV

2

1

2

=

==


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Fig. 3.15: DC Resistance Test (-Connection)

dcs

sdc

dc

s

ss

dc

dc

RR

RR

RR

RR

I

V

2

3

3

2

3

)(2

=

=

==

Exercise 2.5

The results of no-load and blocked rotor tests for a three-phase induction motor are:

o-load test: VL-L = 220V

Pin = 1000W

IL = 20A

P = 400W

Blocked rotor test: VL-L = 30V

Pin = 1500W

IL = 50A

Calculate the per-phase parameters of the approximate equivalent circuit by assuming the

stator are connected in Y.


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3.9 EFFICIECY

Efficiency for any electrical machine is defined as:

+

==

in

in

in

out

P

lossesTotalP

P

P (3.8)

One thing should be noted that theout

P is actually the output from the rotor or motor itself

while the input power comes from the stator.

3.10 TORQUE EQUATIO

Torque equation can be derived from the power equation that is expressed in mechanical

formula and electrical formula. These two formula can be equated together to obtain its

relationship in terms of circuit parameters.

Basic power equation is given as;

TP= where60

2= (speed in rad/sec)

= Speed in rev/min (rpm)

=T Torque in Nm

Then torque,

2

P60T;Torque

= . (3.9)

Eqn. 3.9 is the general formula for torque equation. This formula can be employed to

calculate the output and mechanical torque by some formula modification.

For mechanical torque :2

P60T@T mmmech

= .. (3.10)

For output torque :2

P60T outo

= .. (3.11)


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3.10.1 MECHAICAL TORQUE ( mT )The mechanical torque is sometimes called the induced torque. The mechanical torque

can be also expressed in terms of circuit parameters.

( ) ( ) mr

mr22

2m T

60

T2s1

s

RI3P

===

where =r

speed of the rotor and

=r

The rotor speed in rad/s

( ) ( )( ) ( )

( )( )

s

2

2

2

s

2

2

2

r

22

2

ms

RI3s1ss1RI3

s1s

RI3

T

=

=

=

But,( )

( )121

2

phs

2

XXjRs

R

EI

++

+

= ; so ,( )

( )22

1

2

phs

2

XRs

R

EI

+

+

=

Substitute ( )

( )22

1

2

phs2

XRs

R

EI

+

+

= , into the mechanical torque, mT equation,

Thus,

( )

( )( )[ ]

( )

+

+

=

+

+

=2

2

1

2

s

2

2

phs

s

2

2

2

2

1

2

phs

m

XRs

R

s

RE3

s

R

XRs

R

E3

T

.. (3.12)

By simplifying the above equation, therefore, the formula for the mechanical torque, mT is

( )[ ]( ) ( )[ ]22

12

2

s

2

phs

msXsRR

sRE3T

++==

.. (3.13)

where21

XXX +=


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Fig. 3.16: Motor Torque vs Slip(Speed)

Eqn. 3.13, if we draw on themT versus s on the graph will be appeared as in Fig. 3.16.

From the torque-speed characteristics (Fig. 3.16.) it is observed that;

stT : The torque required by the motor to start. Also called as initial torque.

maxT : The max torque for the motor. Also called as stalling or pull-out torque.

maxS : The slip at

maxT

LT : No-load torque.

FLT : Full-load torque.

s : Synchronous speed.


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3.10.2 MAXIMUM TORQUE ( maxT )

To obtainmaxT , differentiate the mT to obtain maxs ,

From the same curve, the max point can be obtained by differentiating Eqn. 3.13.

0ds

dTm = :

To obtain max torque, then, yield:

( ) 221

2

max

XR

Rs

+

=

Substitutemax

s into Eqn. 3.13, we get:

( )[ ]

( ) ( )( )[ ]2211

s

2

phs

max

XRR

1

2

E3T

++==

(3.14)

3.10.3 STARTIG TORQUE ( stT )

Starting torque can be derived from Eqn. 3.13 with slip, 0.1s =

At starting, 0r= , therefore, 1=

=

s

rss

As a result, the equation of the starting torque, stT is

( )[ ]( )[ ]22

12

2

s

2

phs

stXRR

RE3T

++==

(3.15)

Fig. 3.17 represents the relationship between torque and slip/speed with varyingR .


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3.11.1 USIG PRIMARY RESISTORS

The purpose is to apply a reduced voltage across the motor terminals so that the initial

current is reduced. However, it should be noted that EI and 2ET .

By using the primary resistors, the applied voltage per phase can be reduced by factor of

x.

scst xIT = and scst IxT 2=

ff

f

sc

f

st

f

f

sc

f

f

st

f

st

saxsI

Ix

T

T

sI

xIs

I

I

T

T

22

2

2

22

=

=

=

=

This method is useful for smooth starting small machine. The circuit used for this type of

method is shown below.

Fig. 3.18: Starting of Induction motor using primary resistors

3.11.2 USIG STAR-DELTA STARTER

This method is used for delta-connected motors. It consists of 2-way switch, which

connects the motor in star for starting and delta for normal running, as shown in figure

below.


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At starting, when star-connected, the voltage is reduced by3

1. Hence, the torque

developed is reduced by3

1. This method is cheap and effective provide the starting

torque required does not exceed 1.5 full-load torque. This method is used for machine

tools, pumps and motor-generators.

Fig. 3.19: Starting of Induction motor using star-delta starter

3.11.3 USIG AUTO TRASFORMERThis method can be both for star and delta connected motors. At starting, a reduced

voltage is applied across the motor terminals. When the speed is about 80%, the

autotransformer is cut-off and full supply voltage is applied. The circuit used for this type

of method is shown in figure below.


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Fig. 3.19: Starting of Induction motor using autotransformer

Let the tapping of the transformation ratio = k

ff

fL

stf

fL

st

fL

st saksI

Iks

I

I

T

T 222

2

2

=

=

=

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CHAPTER 3 INDUCTION MOTOR.pdf