Unit-3. Three Phase Induction Motor. - pnbalamuruganpnbalamurugan.yolasite.com/resources/U3 Three...

Unit-3. Three Phase Induction Motor.

N.Balamurugan AP. EEE

UNIT – III

THREE PHASE INDUCTION MOTOR

THREE PHASE INDUCTION MOTOR

Constructional details – Types of rotors – Principle of operation – Slip –

Equivalent circuit – Slip-torque characteristics - Condition for maximum torque

– Losses and efficiency – Load test - No load and blocked rotor tests - Circle

diagram – Separation of no load losses – Double cage rotors – Induction

generator – Synchronous induction motor.

Construction and working principle of three phase induction

motor.

Construction

An induction motor has two main parts. There are,

Stator.

Rotor.

Stator.

Constructed by laminated silicon steel plates by stamping for the purpose of

reducing the Eddy current losses and Hysteresis losses.

Stator has inward projected stator poles; if we increase the no of poles then

speed will be decreased.

It consists stator windings, which are placed in stator slots. Three types of

slots are shown here,



No of poles and types of slots are chosen corresponding to our requirements.

Normally number of poles P = 2*n.

Where, n = number of stator slots/pole/phase.

Stator winding is wounded in two types, are as follows,

* Lap winding.

* Wave winding.

Rotor.

Two basic design types of the rotor are as follows,

o squirrel-cage

o slip ring

Squirrel-cage

Squirrel-cage rotor consist of copper bars slightly longer than the rotor,

which are pushed into the rotor



The ends are welded to copper end rings, so that all the bars are short

circuited permanently. In this type, we can‟t add any more external

resistance to improve the starting torque. So to improve the T st, “Skewed

rotor” has introduced.

In small machines bars and end rings are die cast in alluminium

Skewed rotor.

Slip Ring



It is usually for large 3 phase induction motors.

Rotor has winding, same as stator and the end of each phase is connected

to a slip ring.

In phase wounded rotor always double layer, distributed windings are used

in alternators even the stator has two phases.

The three phase winding stared internally and other three winding

terminals are brought out and connected to three insulated sliprings, which

are mounted on the shaft with brushes resting on them.

Yoke

Made by Alloy cast iron steel.

It provides mechanical support to all parts of the motor and restricts the

magnetic leakage.

Shaft and Bearings

Shaft - cylindrical, solid core type.

Bearings – Ball bearings and roller bearings are used.

Airgap

Should design uniform airgap between rotor and stator.

Working principle.

Rotating Magnetic Field

Balanced three phase windings, i.e. mechanically displaced 120 degrees

from each other, fed by balanced three phase source

A rotating magnetic field with constant magnitude is produced, rotating with

a speed



Where f is the supply frequency and p is the no. of poles and Ns is called

the synchronous speed in rpm (revolutions per minute)

Principle of operation

The stator is connected to a 3-phase AC power supply.

The number of poles is determined by how many times a phase winding

appears. In this example, each phase winding appears two times. This is a two-

pole stator

When AC voltage is applied to the stator, current flows through the windings.

The magnetic field developed in a phase winding depends on the direction of

current flow through that winding.



Start

It is easier to visualize a magnetic field if a start time is picked when no current

is flowing through one phase.

In the following illustration, for example, a start time has been selected during

which phase A has no current flow, phase B has current flow in a negative

direction and phase C has current flow in a positive direction.

Based on the above chart, B1 and C2 are south poles and B2 and C1 are north

poles. A magnetic field results, as indicated by the arrow.



Time 1

At Time 1 phase C has no current flow, phase A has current flow in a

positive direction and phase B has current flow in a negative direction.

Following the same logic as used for the starting point, windings A1 and B2

are north poles and windings A2 and B1 are south poles.

Time 2

Phase B has no current flow. Although current is decreasing in phase A it is

still flowing in a positive direction. Phase C is now flowing in a negative

direction.

At start it was flowing in a positive direction. Current flow has changed

directions in the phase C windings and the magnetic poles have reversed

polarity.

Advantages

Cheaper, Light weight, High efficiency, Require less maintenance.

Disadvantages

Moderate starting torque.

External resistance can‟t be added, so Tst can‟t be controlled.

Applications

o Generally Conveyer line (belt) drives,

o Roller table, Paper mills,

o Traction, Electric vehicles,

o Elevators, pulleys.

(End of the answer.)



Slip.

Slip is the ratio between the slip speed and synchronous speed.

S =

Where,

Ns = synchronous speed in rpm.

N = actual motor speed in rpm.

Equivalent circuit of three phase induction motor .

Generally three phase induction motor is treated as rotating transformer.

Has two winding named by primary and secondary.

Similarly in induction motor, stator → primary, rotor → secondary.

Equivalent circuit of induction motor.



Parameters

V1 = supply voltage in volts.

E1 = Induced emf in stator/phase (due to self inductance)

E2 = Induced emf in rotor/phase (due to mutual inductance)

R1 = stator resistance in ohms.

R2 = rotor resistance in ohms

X1 =stator reactance

X2r =rotor reactance.

E2r =rotor induced emf at running condition.(E2r)

From the diagram, under no load condition the no load current I0,

I0 = Iw+Iµ

Where,

Iw =Working component which supplied no load losses.

Iµ =Magnetic component which sets up flux at core and airgap.

Ro = No load resistance/phase (represents no load losses)

Ro =

X0 = No load reactance/phase (represents flux setup in the core)

Xo =

X2r = sX2

X2 is constant.

S varies with respect to rotor speed.



So the above equivalent circuit will becomes,

Under running condition,

Rotor current I2r =

=

From this ,

Motor load changes → speed changes.

Speed changes → slip changes.

Slip changes → reactance changes.

Reactance X gets differ with speed which is affect the flux, so far rotor

reactance indicated as variable element.

Equivalent circuit for rotor.

I2r =



From the above expression,

X2 and E2 are constant values.

R2 varies with slip.

Now the variable Resistance,

=

= )

From this R2/s has two parts,

o R2 = Rotor copper loss.

o R2( = RL which is vary with slip. So it is indicated as variable element.



Equivalent circuit referred to stator

K = Transformer ratio.

K =

Rotor parameter referred to as stator,

o E2' = E2/K

o I2r' = I2r/K

(Already we have I2r = )

o X2' =X 2/K2

o R2' = RL/K2 = = R2' (

Approximate equivalent circuit



For simplify the calculation the exciting circuit is transferred to the left of

R1, X1.

The inaccuracy because of this shifting is negligible.

For the further simplification,

R01 = R1+R'2 X01 = X1+X'2

R01 = R1+ X01=X1+

I1 = I0+ I2r'

I1 = (IW+I )

Torque equation of three phase induction motor.

Torque is proportional to flux per pole.



2

2

)2

)2

K = 3 / 2П ns

If , S= 1

)2

(Or refer class notes)

Condition for maximum running torque:

Condition: (dT/dR2) = 0;

R22 = X2

2

Condition for maximum running torque:

Torque equation, )2

Condition: = 0;

Sm = is the slip at which torque is maximum.

Tmax= K2

i. The maximum torque is independent of rotor resistance



ii. Maximum torque is directly proportional to the square of the induced

emf at standstill.

iii. Maximum torque is inversely proportional to the rotor reactance.

Torque – Slip characteristics

It consists of three regions

(1) Stable operating region.

(2) Unstable operating region.

(3) Normal operating region.

Stable region

In stable region, the slip value„s‟ is very small. i.e. the term (2 is very

small as compared to .hence neglecting (2

Torque increases slip also increases, motor speed decreases.

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



Unstable region

When the slip is further increases, the region is unstable.

Here,

Slip increases torque decreases, motor speed increases, this region is called

high slip region

Normal operating region

Motor is continuously operated in this region.

Starting torque, maximum torque or pull load torque, full load torque.

Losses in an Induction Motor.

1. Magnetic losses.

2. Mechanical losses.

3. Electrical losses. → variable losses.

Magnetic losses

Also called as core loss (or) iron loss

These losses occur in stator and rotor core

There are Two types

Hysteresis

Eddy current

Hysteresis losses

Due to alternates change in magnetic field in the stator core

To reduce, select silicon for stator for stator &rotor core

}constant losses



Eddy current losses

Due to the flow of eddy current in stator core

To reduce this loss by using laminated construction

Both these losses depends supply frequency (f)

Rotor losses will be neglected. Because this losses depends (Fr)

Fr=sf. Under normal running condition S<<1. So Fr<<1.so for

Losses will be neglected.

Mechanical losses

Fraction & windage losses.

In induction motor slip speed is very small

So consider mechanical losses=constant

Constant losses= Iron Loss+ Mechanical loss

Electrical losses

Stator copper loss+ rotor copper loss

Load vary-IL also vary

Normally stator copper loss are combine with stator copper loss

Rotor copper loss = 3*i22*R2

Power flow diagram

Input power

Output power in stator i.e. input to the rotor P 2 = P I - PSL

Pcu = 3 I2R2

Mechanical power Pm = P 2 - Pcu



Electric power input

PSL

Pcu

Mechanical power

developed Pml

Output mechanical power

Po = Pm - PmL

Motor efficiency = mechanical output power at shaft / electrical input

power to stator

stator

Stator losses

= stator cu

losses +stator

iron loss

Rotor copper

losses

rotor

load Friction and

Windage losses

(mechanical

losses)

Output of stator = input

to the rotor

to stator



Relationship between Rotor input(P2), Rotor Copper loss (Pcu)

and Gross mechanical power (Pm)

Let,

Tg = Gross torque developed by the motor.

P = Tg * ω

P = T g * (2πN/60) general formula…

Power input to the rotor P2 = T g * ωs ωs = synch. Angular freq.

= T g * (2πN s/60)

Gross mech. Developed power in rotor (P m) ,

Pm = T g * (2πN/60) N = actual rotor speed.

Rotor copper loss (P cu) , Pcu = P 2 – P m

= T g * (2π/60) (N s -N)

= T g * (2π/60) slip speed.

Divide P cu by P 2 ,

P cu/ P 2 =

P cu/ P 2 = S

P cu = S*P2

P 2 = P cu/S

Mechanical power developed in rotor (P m) ,

Pm = Rotor input power – Rotor copper losses.

= P 2 – P cu

Tg * (2π/60) (N s -N)

Tg * (2π/60) N s



= P 2 – S*P 2

P m= P 2 (1– S)

From eq.5 &7,

P cu / P m=

P cu / P m=

From the above experiments,

P2 : P m : P cu = P 2 : P 2*(1-S) : S*P2

P2 : Pm : Pcu = 1: (1-S) : S (This is our require d relatio ns hip.)

Load test on Three Phase Induction Motor.

From this test we can find out the input power, output power, developed

torque, power factor, efficiency and the performance of Induction Motor can be

determined.

S*P 2

P2 (1– S)

S

(1– S)

A

S

1

3

0

0

V

,

1

0

A

,

U

P

F

T

RO

T--

OR

V

S

1

S

2

DRUM



Procedure:

Supply is switched on at no load condition,

One set of reading taken at no load.

Then the load is varied up to 120% of load current and not down the readings.

Formula:

1. Power factor = P i n / √3 * V L*IL

2. Slip =

3. Developed torque = (F 1 - F 2)* 9.81* r in N.M .

Where, r = radius of the breakdrum.

4. Output power = 2πNT/60 in watts.

5. Efficiency = Output power / Input power.

No load and Blocked rotor test.

The parameters of equivalent circuit and different losses of a poly phase

Induction Motor can be determined from this test.

By using these tests, we can construct the circle diagram.

NS - N

NS

P0 (Watts)

N (rpm)

T (Nm)

IL (A)

P.f.

% η

Po Vs P.f.

Po Vs % η

Po Vs T

Po Vs IL

Po Vs N

%Slip

Po (W)

T (Nm)

%Slip Vs T

%Slip Vs Po



No load test:

Formula.

1. Total power drawn WNL = W1+W2 in Watts.

2. Power drawn by the motor / phase = WNL /3 in watts.

3. Power factor cos φo = W0 / V o*I o

4. Working current component I W = I o cos φ o

5. Magnetic current component I µ = I o sin φ o

6. Ro = V o / IW ; Xo = V o / Iµ

Blocked Rotor test:

Separation of no load losses can be determined.

At blocked rotor test above 120% current is applied.

Here input power is equal to stator core losses, stator copper losses and

mechanical losses.

A

rotor

rRO

R

V

S

1

S2

BRAKE

DRUM



1. Stator copper loss PS L = 3I o2R 1 in watts.

2. Stator Core loss = Input power – Stator loss. In watts.

Circle Diagram for a three phase Induction Motor.

A

rotor

rRO

R

V

S

1

S2

BRAKE

DRUM



Construction of Circle Diagram

Conduct No load test and blocked rotor test on the induction motor and

find out the per phase values of no load current I0, short circuit current ISC and

the corresponding phase angles Φ0 and ΦSC. Also find short circuit current ISN

corresponding to normal supply voltage. With this data, the circle diagram can

be drawn as follows.

1. With suitable scale, raw vector OA with length corresponding to I0 at an

angle Φ0 from the vertical axis. Draw a horizontal line AB.

2. Draw OS equal to ISN at an angle ΦSC and join AS.

3. Draw the perpendicular bisector to AS to meet the horizontal line AB at C.

4. With C as centre, draw a portion of circle passing through A and S. This

forms the circle diagram which is the locus of the input current.

5. From point S, draw a vertical line SL to meet the line AB.

6. Divide SL at point K so that SK : KL = rotor resistance : stator resistance.

7. For a given operating point P, draw a vertical line PEFGD as shown. Then,

PE = output power, EF = rotor copper loss, FG = stator copper loss, GD =

constant loss (iron loss + mechanical loss)

8. To find the operating points corresponding to maximum power and

maximum torque, draw tangents to the circle diagram parallel to the output

line and torque line respectively. The points at which these tangents touch

the circle are respectively the maximum power point and maximum torque

point

Efficiency line

1. The output line AS is extended backwards to meet the X-axis at O′.

2. From any convenient point on the extended output line, draw a horizontal

line QT so as to meet the vertical from O′. Divide the line QT into 100 equal

parts.



3. To find the efficiency corresponding to any operating point P,

i. draw a line from O′ to the efficiency line through P to meet the

ii. Efficiency line at T1. Now QT1 is the efficiency

Slip Line

1. Draw line QR parallel to the torque line, meeting the vertical through A at

R. Divide RQ into 100 equal parts.

2. To find the slip corresponding to any operating point P, draw a line from A

to the slip line through P to meet the slip line at R1.

1. Now RR1 is the slip

Power Factor Curve

a. Draw a quadrant of a circle with O as centre and any convenient radius.

Divide OCm into 100 equal parts.

b. To find power factor corresponding to P, extend the line OP to meet the

power factor curve at C′. Draw a horizontal line C′C1 to meet the vertical

axis at C1. Now OC1 represents power factor.

Double cage Induction Motor.



Squirrel cage induction motor has excellent running characteristics, but it‟s

starting torque is low.

Slipring induction motor has very high starting torque and good running

characteristics. But this motor is not suitable for so many applications.

So to eliminate these drawback, double cage induction motor has introduced

with,

Good starting torque.

Excellent running characteristics.

Upper cage.

Made by high resistive materials (Brass, Alluminium.)

It has low leakage reactance.

Inner cage.

Made by low resistive materials (copper.)

It has low leakage reactance.

The cage windings are in parallel

The current distributions of the windings are inversely proportional to the

impedance. So inner cage winding current is very small compare to outer

cage winding current.

In inner cage winding the current flow lags the induced emf by large angle

because of large leakage reactance. So produce high starting torque.

Similarly, in outer cage winding, current flow nearly inphase with induced

emf due to low XL. so produced high starting torque and good running

characteristics.

At starting time the rotor current frequency is equal to the stator current

frequency.



When the motor speed is increases the rotor current frequency starts

decreasing.

At normal speed IR frequency is 2 to 3HZ. At this frequency XL of both

cages are negligible compare with their resistance.

During normal condition IR will be divide between inner & outer cage

winding inversely proportional to their distances.

The resistance of this upper cage winding is 5 to 6 times of the inner cage

winding.

Due to this upper cage current is very low compared to image winding.

Thus during normal operating conditions the torque developed by the

increase winding.

Slip – Torque characteristics.

Equivalent Circuit.

Advantages.

Can be used frequently under heavy load conditions.



IInndduuccttiioonn GGeennee rraattoorr..

The Induction motor‟s rotor is to be driven by another machine at above Ns.

now the induction motor runs as a generator, such a arrangement is called “

Induction Generator “

P↑ ↓Q

Prime mover 3φ I.Motor

When rotor speed > Ns, then slip S becomes –ve.

Consequently rotor emf E2 and rotor current I2 and stator component I2'

change their sign.

The developed mechanical power (Pm) and torque (T) become

–ve because of –ve slip.

When Nr > Ns , the rotor does not supply mechanical power to the shaft but

absorbs mechanical power from the shaft.

This mechanical power converted in to electrical power and this energy is

related by stator.

The induction generator absorbs reactive power from the lines for creating

it‟s own magnetic field.

It is also called “synchronous generator. “

Advantages.

Need very small auxiliary equipments.

Can be run in parallel with generators at any frequency without hunting.

Robust construction.

Speed variation of prime motor is less important.



Disadvantages.

Poor efficiency.

Draws lagging kVAR.

Can operate only at leading power factor.

Synchronous Induction Motor.

A three phase Induction Motor runs at constant speed when the rotor

winding is fed from a DC source. Such motors are called “Synchronous

Induction Motor”

A DC supply is fed to the conventional three phase rotor winding.

Due to this an alternate N & S poles on the rotor, forms with fixed space.

These fixed rotor poles get magnetically locked with rotating magnetic field

produced by three phase stator winding.

Start: The machine is started as an ordinary slip-ring induction motor.

Run: The additional resistances are completely cutout and the motor runs

with a small slip.



Then this motor has both slip-ring induction motor feature (high Tst) and

synchronous motor feature (constant speed).

Advantages.

Possible to start with heavy loads.

Cogging.

The motor cannot start itself due to magnetic locking between rotor and

stator slots.

To overcome this problem by making the number of rotor slots prime to the

number of stator slots.

Crawling.

The motor runs at one seventh of it‟s Ns with low pitched howling sound

is called “Crawling.”



2 Mark Question And Answer

1) State the principle of 3 phase induction motor?

(APIRAL /MAY 2009)

While starting, rotor conductors are stationary and they cut the revolving

magnetic ield and so an emf is induced in them by electromagnetic induction.

This induced emf produces a current if the circuit is closed. This current

opposes the cause by Lenz‟s law and hence the rotor starts revolving in the

same direction as that of the magnetic field.

2) What are the advantages and disadvantages of direct load test for 3 –

Phase I M? (MAY/JUNE- 2006)

Advantages

Direct measurement of input and output parameters yield accurate results

Aside from the usual performance other performances like mechanical

vibration, noise etc can be studied.

By operating the motor at full load for a continuous period, the final steady

temperature can be measured.

Disadvantages

Testing involves large amount of power and the input energy and the entire

energy delivered is wasted

Loading arrangement cannot be provided for motors of large power rating

3) Applications of induction motor? (NOVEMBER/DECEMBER-2009)

(MAY/JUNE- 2009)

Conveyer line (belt) drives,

Roller table, Paper mills,

Traction, Electric vehicles,



Elevators, pulleys,

Air-conditioning and any industrial process that requires variable-speed

operation.

4) List out the method for speed control of 3phase cage type induction motor?

(MAY/JUNE- 2009)

By changing supply frequency

By changing no of poles

By operating the two motors in cascade

By changing supply voltage.

5) State the advantages of skewing? (MAY/JUNE- 2007)

It reduces humming and hence quite running of motor is achieved. It

reduces magnetic locking of the stator and rotor.

6) What are the losses occurring in an I M and on what factors do they

depend? (NOVEMBER/DECEMBER-2007)

o Magnetic losses Wi

o Electrical losses Wcu

o Mechanical losses Wm

For I M operating in normal condition (with constant voltage and frequency)

magnetic and mechanical losses remain constant whereas electrical losses vary

in square proportion to the current.

7) What care should be taken at the time of construction to reduce eddy

current losses in Induction motors? (NOVEMBER/DEC-2010)

Make the resistance of the core body as large as possible. This is achieved by

laminating the stator core, stacked and reverted at right angles to the path of



eddy current. The laminations are insulated from each other by thin coat of

varnish.

8) Why is there a not appreciable magnetic loss in the rotor core of Induction

motors? (MAY/JUNE- 2006)

Although the rotor core is also subjected to magnetic flux reversals and

since the frequency of flux reversals in the rotor, fr = Sfs, is very small, the iron

loss incurred in the rotor core is negligibly small.

9) What is meant by synchronous watt? (NOVEMBER/DEC-2008)

With the power input to the motor Pi, after the losses in the stator

winding. Wcu1 and stator core, Wi, are met with, the remaining power is

transferred to the rotor by the rotating magnetic field as power input to the rotor

Pir,

Pir = Pi –wau1 - Wi

The power input to the rotor Pir is transferred from the stator to the rotor by

rotating magnetic field which rotates at synchronous speed Ns. Torque Td is

developed in the rotor as a result of P ir and the equation for Pir can alternatively

be expressed as

Td = Pir Syn

9) Name the tests to be conducted for predetermining the performance of 3-

phase induction machine. (MAY/JUNE- 2006)

(a) No load test

(b) Blocked rotor test

10) What are the information’s obtained from no-load test in a 3-phase

I M? (NOVEMBER/DECEMBER-2009)

(i) No –load input current per phase, Io



(ii) No load power factor and hence no load phase angle

(iii) Iron and mechanical losses together

(iv) Elements of equivalent circuit shunt branch

11) What is circle diagram of an I M? ̀ (MAY/JUNE- 2005)

When an I M operates on constant voltage and constant frequency source,

the loci of stator current phasor is found to fall on a circle. This circle diagram is

used to predict the performance of the machine at different loading conditions

as well as mode of operation.

12) What are the advantages and disadvantages of circle diagram

method of predetermining the performance of 3 – phase I M?

(NOVEMBER/DECEMBER-2007)

The prediction can be carried out when any of the following information is

available

The input line current. The input power factor, The active power input, The

reactive power input,

The apparent power input, the output power, the slip of operation, the torque

developed,

The equivalent rotor current per phase, Maximum output power, and Maximum

torque developed.

The only disadvantage is, being a geometrical solution, errors made during

measurements will affect the accuracy of the result.

13) Why an induction motor is called as rotating transformer?

(MAY/JUNE- 2008)

The rotor receives same electrical power in exactly the same way as the

secondary of a two winding transformer receiving its power from primary. That

is why induction motor is called as rotating transformer.



14) Why an induction motor never runs at its synchronous speed?

(NOVEMBER/DECEMBER-2009)

If it runs at synchronous speed then there would be no relative speed

between the two, hence no rotor emf, so no rotor current, then no rotor torque to

maintain rotation.

15) What are slip rings? (MAY/JUNE- 2009)

The slip rings are made of copper alloys and are fixed aroud the shaft

insulating it. Through these slip rings and brushes rotor winding can be

connected to external circuit.

16) What is the advantage of cage motor? (MAY/JUNE- 2006)

Since the rotor has low resistance, the copper loss is low and efficiency is

very high. On account of simple construction of rotor it is mechanically robust,

initial cost is less, maintenance cost is less, simple starting arrangement.

Unit-3. Three Phase Induction Motor. - pnbalamuruganpnbalamurugan.yolasite.com/resources/U3 Three...

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