ele290 chapter 6

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BASIC ELECTRICAL ENGINEERING ELE 290

6.0 INDUCTION MOTORINDUCTION MOTOR

6.1INTRODUCTION TO INDUCTION 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 that manner. 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 the secondary of 2winding transformer.

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

Most appliances, such as washing machines and refrigerators, use a single-phase inductionmachine

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

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

6.2 CONSTRUCTION

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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6.2.1 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. 6.0: 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.

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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.

Fig. 6.1: Wound Rotor

The rotor is cylindrical and is made up of a three phase windings with terminals brought outto 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

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6.2.2 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. 7.2. 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 the stator

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.

Fig. 6.2: A Typical Stator

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6.3 PRODUCTION OF ROTATING MAGNETIC 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 speedsN . The flux is called as rotating magnetization field. The

mathematical equation is given as:

p

f120Ns 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. 6.3.

Fig. 6.3: 3-Phase Current

tsinitimR

120tsinitimY

120tsinitimB

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6.4 PRINCIPLE OF OPERATION

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. 6.4.

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. 6.5.

Fig. 6.4: Conductor A is located under north pole and conductor B is located under south pole

Fig. 6.5: Right Hand Fleming Fig. 6.6: Ampere Right Hand Rule

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The same process happens to conductor B.

As shown in Fig. 6.6, 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. 6.7.

Fig. 6.7: 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 at sN .

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6.5 SLIP AND SYNCHRONOUS SPEED

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

speed:

s

rs

N

NNs

.. (3.0)

where: sN synchronous speed in rev/min.

r

N rotor speed in rev/min.

From Eqn. 3.0, the rotor speed can be derived as s1NNsr . 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,rf 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 speed rs NN

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6.6 PER PHASE EQUIVALENT CIRCUIT

The 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. 6.8 below.

Fig. 6.8: Per-Phase Equivalent of 3-Phase Induction Motor.

The per-phase equivalent circuit referred to the stator winding is shown in Figure 7.9. This equivalent

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

circuit.

Fig. 6.9: Per-Phase Equivalent Circuit Referred to Stator Winding.

From the equivalent circuit;

1I Stator phase current. : 2o1 III

= Stator line current (for stator Y-connection)

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=3

IL , where

LI stator line current (for stator: - connection)

2I Rotor current referred to stator winding

oI No-load current,

mco III

cI Core current

mI Magnetizing current

For approximate equivalent circuit;

2o1 III

mco III

;

m

phs

m jX

EI ;

121

2

phs

2

XXjRsR

EI

... (3.1)

This model is normally used for analysis purposes for simplicity.

c

phs

c R

EI

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6.7 POWER FLOW DIAGRAM

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

Input power = Output Power + Losses

Core Loss Iron Loss

Stator Input Power = Stator Output Power + Stator Losses

Stator Iron Losses ( SIL ) @Core Loss (Pcore)

( Normally Given )

Stator Cooper Losses ( SCL )@ PSCL

Rotor Input Power ( RIP) @

Air-gap Power

Fig. 6.10: Power Flow Diagram

Fig. 6.10 shows the power flow diagram while Fig. 7.11 illustrates the components that involve in the

power losses calculation.

Fig. 6.11: Components Involved in Power Flow Diagram

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From the circuit shown in Fig. 6.11, the power equations can be derived as follows:

[i] SCL occurs atsR , then

s

2

2 RI3SCL . (3.2)

[ii] RCL occurs at R2 'RR , then 22

2 RI3RCL . (3.3)

[iii]mP occurs at s1

s

R2

, then s1s

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 2

2

2m =

s

s1RCL = s1RIP (3.6)

From Eqn. 3.5 : s

RI3RIP 2

2

2 =

s1Pm

=s

RCL (3.7)

From these equations, we do not need to recalculate mP andRIP, if RCL is known provided that

the value for slip is known.

The input power comes from the stator input, then:

cosIV3cosIV3PphphLLin

WhereLV

andLI

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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6.8 EFFICIENCY

Efficiency for any electrical machine is defined as:

losses.totalP

P

P

P

out

out

in

out

(3.8)

One thing should be noted that theoutP is actually the output from the rotor or motor itself while the

input power comes from the stator.

6.9 TORQUE EQUATION

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 where 60

N2 (speed in rad/sec)

N Speed in rev/min (rpm)

T Torque in Nm

Then torque,

N2

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 :N2

P60T@T mmmech

.. (3.10)

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For output torque :N2

P60T outo

.. (3.11)

6.9.1 MECHANICAL 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

TN2s1

s

RI3P

where r

N speed of the rotor and

r

The rotor speed in rad/s

s

2

2

2

s

2

2

2

r

22

2

ms

RI3

s1s

s1RI3s1

s

RI3

T

But,

121

2

phs

2

XXjRs

R

EI

; so ,

22

1

2

phs

2

XRs

R

EI

Substitute

22

1

2

phs

2

XRs

R

E

I

, into the mechanical torque, mT equation,

Thus,

2

2

12

s

2

2

phs

s

2

2

2

2

1

2

phs

m

XRsRs

RE3

s

R

XRs

R

E3

T

.. (3.12)

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

2212

2

s

2

phs

msXsRR

sRE3T

.. (3.13)

where21

XXX

Fig. 6.12: Motor Torque vs Slip(Speed)

Eqn. 3.13, if we draw on the mT versus s on the graph will be appeared as in Fig. 6.12. From the

torque-speed characteristics (Fig. 6.12.) 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

NLT : No-load torque.

FLT : Full-load torque.

sN : Synchronous speed.

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6.9.2 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

Substitutemaxs into Eqn. 3.13, we get:

22

11s

2

phs

max

XRR

1

2

E3

T (3.14)

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6.9.3 STARTING TORQUE ( stT )

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

At starting, 0Nr , therefore, 1

Ns

NrNss

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

22

12

2

s

2

phs

stXRR

RE3T

(3.15)

Fig. 7.13 represents the relationship between torque and slip/speed with varyingR .

Fig. 6.13: Motor Torque vs SIip(Speed)

The relationship between stT and maxT can be obtained by assuming stator resistance =0. From Fig.

6.13, we found that at starting, torque is proportional to the resistance.

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Tutorial

QUESTION 1

A 15 hp, 415V, 50Hz, 4 pole, three-phase star connected induction motor operates at 1465 rpm when carrying

rated load. The motor parameters refer to the stator express ed in ohm per phase are:

3.0Rs 83.0Xs 31.0'Rr 25.1'Xr 31Xm

The total rotational losses are 1100 W and are assumed to be constant. At rated voltage and rated frequency,

find:

i) The motor slipii) The stator currentiii) The stator copper l ossiv) The power converted fro m electrical to mechanical formv) The overall machines efficiency

QUESTION 2

A si ngle phas e transformer has the following circuit parameters :

20KVA, 1500/5000V, 50Hz, 0.95 lagging power factor.

R1= 0.5 R2= 1.5

X1= 2.0 X2= 0.45

Rc= 350 Xm= 500

Calculate;

i) The approximate equivalent ci rcuit parameter referred to pri mary side.ii) The full -load current on the primary and secondary windings.iii) Assume that this transformer is supplying rated l oad at 5000V and 0.95 la gging power factor.

What is the trans former i nput voltage?

iv) What is the trans former efficiency under condition of part (3).v) What does the power losses essentially show for the short circuit and open circuit tests.

QUESTION 3

Draw and label the related power equation for the power flow di agram of the induction motor.

QUESTION 4

A three-phase 5hp, 50 Hz, 4 poles induction motor operating at rated voltage and frequency with rated load

has an efficiency of 84.5%. The stator conductor l oss , rotor conductor loss and core loss are 225 W, 153 W, and

115 W respectively. Find the percent of sl ip and the shaft speed for this induction motor.

A 3-phase 415V, 50Hz, 4 poles induction motor running at 1450rpm. The motor is taking 70kW at 0.85 lagging

power factor. The total s tator losses are 3.5KW and the friction mechanical losses are 1.5kW. Determine:

i) The motor input current

ii) The motor slip

iii) The rotor copper loss

v) The overall motor efficiency

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QUESTION 5

A 415 V, three-phase, 8-pole, 50 Hz, star-connected induction motor runs at 5% slip. Fi nd:

i) The speed of the rotational magnetic field.ii) The speed of the rotor.iii) The sl ip-speed of the rotor.iv)

The frequency of the i nduced emf in the rotor ci rcuit.

QUESTION 6

The motor s tated in Question 1 has an output power of 20 hp at the shaft when operating on full -load when

the slip is 0,05. Its friction and windage losses are found to be 500 W, Determine;

i) The shaft power in Watts.ii) The shaft torque.iii) The mechani cal power developed in the rotor.iv) The induced torque,v) The air gap power.

QUESTION 7

A three phase, star-connected 440 V, 50 Hz s ix -pole i nduction motor is rated at 100 hp. The equivalent ci rcuit

parameters

Rs = 0.084 Rr = 0.066 Xm = 6.9

Xs = o.2 Xr' = 0.165

Protor=1.5kW Pcore=1.0kW

For a sli p of 0.035, find

i) The line current I1ii) The stator copper l ossesiii)

The ai r gap poweriv) The power converted fro m electrical to mechanical form

v) The induced torquevi) The load torquevii) The overall machine efficiency

QUESTION 8

Prove that the speed of an induction motor Nr i s given as:

Nr = Ns(1-s)

Where Ns = synchronous speed.

s= slip

QUESTION 9

State two methods of starting an induction motor.

QUESTION 10

A 3-phase 440V, 60Hz, 4 pol es induction motor running at 1710rpm. The motor i s

taking 60kW at 0.75 lagging power factor. The total stator losses are 2.5KW and the

friction and windage tosses are 2.0kW. Determine:

i) The motor input currentii) The motor slipiii)

The rotor copper los s

iv) The brake horse powerv) The overall motor efficiency

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QUESTION 11

A 440V, 60Hz, si x pol es Y-connected induction motor is rated at 20hp. Its equivalent circuit components are:

Rs = 0.344 Xs = 0.498 Xm = 12.4

Rr1 = 0.145 Xr' = 0.226

The rotational losses including the core losses amount to 260W. The s lip of the machi ne is 0.03. Draw the

induction machine approximate equivalent ci rcuit.

Determine:

i) The induced electromechanical power.ii) The output power.iii) The output torque.

QUESTION 12

A 3-phase, 415V, 50Hz, 6 poles, wye-connected induction motor has the following parameters:-

R1= 0.06 R2 = 0.24 Xm = 7.5

X1 = 0.35 X2 = 0.21

PF&W = 2.7kW PMISC = 230W

Under load, the motor operates at 950rpm. Using approximate equivalent ci rcuit, determine:-

i) stator line currentii) stator power factoriii) the developed torqueiv) the overall efficiencyif the induction motor connection is rearranged to operate using delta connection, determine the

stator line current

ele290 chapter 6

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