Unit 3 - Induction Motor PART - A 1. 2. - Rajkumar Kuppapillai · PDF fileUnit 3 - Induction...

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K.Rajkumar, Associate Professor/EEE, Dhanalakshmi Srinivasan Institute of Technology, Samayapuram.

Unit 3 - Induction Motor

PART - A

1. Why the rotor slots are slightly skewed in squirrel cage IM? (or)Why are the clots on the

cage rotor of induction motor usually skewed?

The rotor bars are not parallel to the shaft but slightly skewed for the following purposes,

i. It helps to make the motor run quietly by reducing magnetic hum and

ii. It avoids the tendency of magnetic locking between stator and rotor teeth.

iii. It gives more uniform torque in running.

2. A 50 Hz, 6 pole, 3–phase induction motor runs at 970 rpm. Find slip.

Rotor Speed, Nr=970 r.p.m

Synchronous Speed, mprp

fN s ..1000

6

50120120

%)(., 30301000

9701000or

N

NNsSlip

s

rs

3. Under what condition, the slip in an induction motor is

(a) Negative

(b) Greater than one.

The slip of the induction motor is negative when it is operated above the rated speed (as

induction generator).

The slip is greater than one when the motor is operated at brake mode.

4. What are the two fundamental characteristics of a rotating magnetic field?

i. The resultant flux is constant i.e, 1.5 times the maximum value of the flux due to any

phase.

ii. The resultant flux rotates around the stator at synchronous speed given by ,

mprp

fN s ..

120

5. Define slip of an induction motor.

The difference between the synchronous speed Ns and the actual speed Nr of the rotor is

known as slip.

100

s

rs

N

NNSlip%

6. What are the merits and demerits of double squirrel cage induction motor?

Merits (or) Advantages:

i. Higher starting torque

ii. Wide range of torque-speed characteristics can be obtained

Demerits (or) Disadvantages:

i. Costlier about 20-30% of single cage rotor

ii. Effective rotor resistance is high, so rotor heating at the starting is more.

iii. Due to higher effective leakage reactance, the powerfactor is reduced.

2


7. How do change in supply voltage and frequency affect the performance of a 3 phase

induction motor?

The torque developed by the induction motor is proportional the voltage applied ( T α E2)

The speed of the motor is inversely proportional to the supply frequency . (Ns α 1/f).

8. The starting torque of a squirrel cage induction motor cannot be altered when the applied

voltage is constant. Why?

22

22

222

XR

RkETst

From the above relation, R2 is constant and X2 is independent of‘s’. So, Tst is only depending

on E22. Therefore the starting torque cannot be altered for constant supply voltage, where

E2αV.

9. A 3-Phase, 4 pole, induction motor operates from a supply whose frequency is 50Hz.

Calculate the frequency of the rotor current at standstill and the speed at which the

magnetic field of the stator is rotating.

sff' Frequency, Current Rotor . At standstill, s=1. Therefore, f’=50Hz.

Speed of rotation of magnetic field, r.p.m 1500p

fN s

4

50120120

10. Write down the condition to get maximum torque under running condition.

The condition for maximum torque of 3 phase induction motor is, 22 sXR 11. Define Cogging.

With the equal number of stator and rotor slots, the speeds of all the harmonics produced by

the stator slotting coincide with the speed of rotor harmonics. Thus the machine refuses to

start due to magnetic locking. This is known as cogging.

Therefore the number of stator slots should not be equal to the number of rotor slots.

12. Define Crawling.

The tendency of the motor to run stably at speeds as low as one-seventh of its synchronous

speed is called crawling. The torque produced during this crawling is a harmonic induction

torque.

13. How the slip does vary with load?

The greater the load, the greater the slip.

14. What are the advantages of double cage induction motor?

i. Starting torque is high

ii. Low starting current hence suitable for direct on line starting

iii. Wide range of torque-slip characteristics can be obtained

15. What is an Induction generator? What are its limitations?

When a 3-phase induction motor made to run at a speed higher than its synchronous speed by

means of prime mover, it becomes 3-phase induction generator.

Limitations:

i. It needs reactive power from the supply network.

ii. Its output voltage and frequency cannot be controlled.

iii. It always operates at a leading powerfactor.

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PART – B

1. a. Explain the principle of operation of 3-phase induction motor and explain how the

rotating magnetic field is produced by three-phase currents. (10)

Principle of Operation:

Consider a portion of 3-phase induction motor as shown in Fig. (8.13).

The operation of the motor can be explained as follows:

i. When 3-phase stator winding is energized from a 3-phase supply, a rotating magnetic

field is set up which rotates round the stator at synchronous speed Ns(=120f/p).

ii. The rotating field passes through the air gap and cuts the rotor conductors, which as yet,

are stationary. Due to the relative motion between the rotating flux and the stationary

rotor, e.m.f’s are induced in the rotor conductors. Since the rotor circuit is short-circuited,

currents start flowing in the rotor conductors.

iii. The current-carrying rotor conductors are placed in the magnetic field produced by the

stator. Consequently, mechanical force acts on the rotor conductors. The sum of the

mechanical forces on all the rotor conductors produce a torque which tends to move the

rotor in the same direction as the rotating field.

iv. The fact that rotor is urged to follow the stator field (i.e., rotor moves in the direction of

stator field) can be explained by Lenz’s law. According to this law, the direction of rotor

currents will be such that they tend to oppose the cause producing them. Now, the cause

producing the rotor currents is the relative speed between the rotating field and the

stationary rotor conductors. Hence to reduce this

relative speed, the rotor starts running in the same

direction as that of stator field and tries to catch

it.

Rotating Magnetic Field (RMF):

Figure 1 shows three-phase windings displaced in

space by 120º, are fed by three-phase currents, displaced

in time by 120º, they produce a resultant magnetic flux,

which rotates in space as if actual magnetic poles were

being rotated mechanically.

The sinusoidal flux due to three-phase windings is

shown in Figure 2. Let the maximum value of flux due to

any one of the three phases be Φm. The resultant flux Φr, Figure 1 Three Phase Winding in motor

4


at any instant, is given by the vector sum of the individual fluxes, Φ1, Φ2 and Φ3 due to three

phases. We will consider values of Φr at four instants 1/6th

time-period apart corresponding to

points marked 0, 1, 2 and 3 in Figure 2.

Figure 2 Three Phase Rotating Sinusoidal Fluxes and their 120odisplaced vector

Referring to Figure 2,

a. At =0o,

mm and 2

3

2

30 321 ,

The flux Φ2 is drawn in a direction opposite to the direction

assumed positive. Therefore,

mm

o

mr 2

3

2

33

2

60

2

32 cos

b. At =60o and Point 1 in Figure 2,

02

3

2

3321 andmm ,

Therefore,

mm

o

mr 2

3

2

33

2

60

2

32 cos

It is clear that the resultant flux is again the same and rotated in

clockwise direction through an angle of 60o.

c. At =120o and Point 2 in Figure 2,

mm and 2

30

2

3321 ,

Therefore,

mm

o

mr 2

3

2

33

2

60

2

32 cos

It is found that the resultant flux is again the same and rotated in

clockwise direction through an angle of 60o.

d. At =180o and Point 3 in Figure 2,

mm and 2

3

2

30 321 ,

The resultant flux, mr 2

3 and has rotated clockwise through an additional angle 60

o.

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The important characteristics of the rotating magnetic field,

i. The resultant flux is of constant value

mr

2

3 due to any

phase.

ii. The resultant flux rotates around the stator at synchronous

speed given by p

fNs

120

b. Discuss the different power stages of an induction motor with losses. (6)

The electric power input given to the stator of an induction motor is converted into

mechanical power output at the shaft of the motor. The various losses during the energy

conversion are:

1) Fixed Losses:

a. Stator Iron Loss

b. Friction and Windage Loss

The rotor iron loss is negligible because the frequency of the rotor current is very small

under running condition.

2) Variable Losses

a. Stator Copper Loss

b. Rotor Copper Loss

The above figure shows that the electric power fed to the stator suffers losses and finally

converted into mechanical power.

The important points to be noted from the diagram are,

i. The stator input Pin= Stator output (Rotor Input) P2+ Stator I2R Loss+Stator Iron Loss

ii. Rotor input P2= Stator Output

It is because stator output is entirely transferred to the rotor through the airgap by

electromagnetic induction.

iii. Mechanical Power Developed, Pm=Pg-Rotor Copper Loss.

This mechanical power available is the gross rotor output and will produce a gross torque

Tg.

iv. Mechanical shaft output, Psh= Pm- Friction and Windage Loss.

Mechanical power available at the shaft produces a shaft torque Tsh.

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2. a. Derive an expression for the torque of an induction motor and obtain the condition for

maximum torque. (8)

The torque developed is proportional to the armature current, flux per pole and the

powerfactor.

22 cosIT or 22 cosIkT

where,

I2=Rotor current at standstill

ϕ2= Angle between rotor e.m.f and rotor current

k=Constant

At standstill, the rotor e.m.f E2 is proportional to the flux ϕ.

Therefore, 222 cosIkET

Here, 2

222

2

2

2

22

XsR

Es

Z

sEI

and

22

222

2

2

2

XsR

R

Z

R

cos

Then the torque, 2

222

2

2

22

222

22

XsR

R

XsR

EskET

22

222

222

XsR

REksT

--- (1)

At the maximum torque, 02

dR

dT.

0

22

222

222

22

XsR

REks

dR

d

dR

dT

02

222

222

2222

22

22

222

XsR

RREksEksXsR

02

222

222

222

222

222

XsR

EksRXsR

02 22

22

222 RXsR

22

22

2 RXs or 22 RsX or 2

2

X

Rs max ---- (2)

Therefore the maximum torque is obtained by substituting the slip at maximum torque, smax.

222

22

22

22

2

2

222

222

2

2

22

222

222

2 XR

REk

XX

RR

REX

Rk

XsR

REksT

max

maxmax

2

22

2X

EkT max ---- (3)

From the equation (3), it is concluded that,

The maximum torque is independent of rotor resistance

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The maximum torque is obtained when the rotor reactance equals its resistance.

Maximum torque varies inversely as standstill reactance. Hence, it should be kept as

small as possible.

Maximum torque varies directly as square of the applied voltage.

b. Draw and explain the slip-torque characteristics of a typical 3-phase induction motor.

Mark the starting and maximum torque regions on the diagram so drawn. (8)

The motor torque under running conditions is given by,

2

222

2

222

XsR

REksT

A curve drawn between the torque and slip, for a particular value of rotor

resistance R2 is called torque-slip characteristic. Figure shown below gives a set of torque-slip

characteristics for a slip-range from 0 to 1 for various value of rotor resistance.

Figure 3. Torque - Slip Characteristics

From the Figure 3, the following points to be noted,

i. At slip, s = 0, torque, T = 0 so that torque-slip curve starts from the origin.

ii. At normal speed, slip is small so that sX2 is negligible as compared to R2. Then the torque

will be proportional to, s/R2.

i.e, 2R

sT and finally sT when R2 is constant.

Hence torque slip curve is a straight line from zero slip to a slip that corresponds to full-

load.

iii. As slip increases beyond full-load slip, the torque increases and becomes

maximum at smax=R2/X2. This maximum torque in an induction motor is called pull-out

torque or break-down torque. Its value is at least twice the full-load value when the

motor is operated at rated voltage and frequency.

iv. As the slip increases further with increased load, then R2 becomes negligible compared to

sX2. Therefore, for larger value of slip, 22

2 Xs

sT or

sT

1 . Hence, the torque-slip

characteristic is a rectangular hyperbola. It is important to understand that the load

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increased further decrease the torque beyond the maximum torque and eventually stops at

one point.

v. It is also seen that the maximum torque does not depend on R2, but it decides the value of

slip at which the maximum torque occurs.

3. a. Describe the construction and principle of operation of a 3-Phase induction motor with

neat sketch. (10)

A 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotor

is separated from the stator by a small air-gap which ranges from 0.4 mm to 4

mm, depending on the power of the motor.

Stator:

It consists of a steel frame which encloses a hollow, cylindrical core

made up of thin laminations of silicon steel to reduce hysteresis and

eddy current losses. A number of evenly spaced slots are provided

on the inner periphery of the laminations .The insulated connected

to form a balanced 3-phase star or delta connected circuit. The 3-

phase stator winding is wound for a definite number of poles as per

requirement of speed. Greater the number of poles, lesser is the

speed of the motor and vice-versa. When 3-phase supply is given to

the stator winding, a rotating magnetic field of constant magnitude

is produced. This rotating field induces currents in the rotor by

electromagnetic induction.

Rotor:

The rotor, mounted on a shaft, is a hollow laminated core

having slots on its outer periphery. The winding placed in

these slots (called rotor winding) may be one of the

following two types: (i) Squirrel cage type (ii) Wound

type

i. Squirrel cage rotor:

It consists of a laminated cylindrical core having

parallel slots on its outer periphery. One copper or

aluminum bar is placed in each slot. All these bars are

joined at each end by metal rings called end rings as

shown in Figure 5. This forms a permanently short-

circuited winding which is permanent. The entire

construction (bars and end rings) resembles a

squirrel cage and hence the name. The rotor is not

connected electrically to the supply but has current

induced in it by transformer action from the stator.

Those induction motors which employ squirrel cage

rotor are called squirrel cage induction motors. Most

of 3-phase induction motors use squirrel cage rotor

as it has a remarkably simple and robust construction

enabling it to operate in the most adverse circumstances. However, it suffers from the

disadvantage of a low starting torque. It is because the rotor bars are permanently short-circuited

Figure 4 Stator

Figure 5 Squirrel Cage Rotor

Figure 6 Slip Ring Rotor

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and it is not possible to add any external resistance to the rotor circuit to have a large starting

torque.

ii. Wound rotor:

It consists of a laminated cylindrical core and carries a 3- phase winding, similar to the one

on the stator as shown in Figure 6. The rotor winding is uniformly distributed in the slots and is

usually star-connected. The open ends of the rotor winding are brought out and joined to three

insulated slip rings mounted on the rotor shaft with one brush resting on each slip ring. The three

brushes are connected to a 3-phase star-connected rheostat. At starting, external resistances are

included in the rotor circuit to give a large starting torque. These resistances are gradually

reduced to zero as the motor runs up to speed.

b. Deduce and discuss the equivalent circuit of 3-phase induction motor.(6)

The equivalent circuit of 3-phase induction motor can be developed on per phase basis by

following the same procedure as used for the development of equivalent circuit of transformer.

Let,

V1=Applied voltage per phase to the stator winding

I1 = Stator current per phase

R1 = Stator winding resistance per phase

X1 = Leakage reactance of the stator winding per phase

E1=Induced EMF per phase in stator winding

=4.44fϕT1Kw1=V1-I1(R1+jX1)

I0=No-load current per phase = Iw+Im

Im =Magnetizing current per phase

Iw= Loss component of no-load current per phase

E2= Induced EMF per phase in rotor winding at standstill

=4.44fϕT2Kw2=I2(R2+jX2)

I2 = Rotor current per phase

R2 = Rotor winding resistance per phase

X2 = Leakage reactance of the rotor winding per phase at standstill

When the motor is running at a slip, ‘s’ then the emf induced in the rotor is sE2 at slip frequency.

Hence the rotor has been represented by a resistance R2 and a variable leakage reactance sX2 (due

to slip frequency) with variable rotor voltage sE2.

Figure 7 Per phase Equivalent Circuit of 3-phase induction motor

Referring to Figure 7, the rotor current per phase is given by,

22

2

22

22

jXs

R

E

jsXR

sEI

--- (1)

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Based on the equation (1), the rotor circuit modified to represent the rotor resistance with respect

to the change in slip as R2/s. This modification is done so that the emf induced in the rotor is

independent of slip and it becomes easier to refer the rotor quantities to stator.

Figure 8 Modified Equivalent Circuit per phase

Self induced e.m.f E1 is induced in the stator winding and mutually induced e.m.f

ratio tiontransforma the is K wheresKEsEE ,'122 is induced in the rotor winding. The

secondary values may be transferred to the primary and vice versa. As in the case of transformer,

it should be remembered that when shifting impedance or resistance from secondary to primary,

it should be divided by K2 whereas current should be multiplied by K.

Rewriting the equation (1),

22

2

22

2 1

jXs

R

E

KK

I

2

222

22

KjXKs

R

E

KK

I

22

22

22

KjXKs

R

EK

K

I

''

'

22

12

jXs

R

EI

Figure 9 Per phase equivalent circuit of 3-phase induction motor referred to stator

The equivalent circuit can be further modified so that the hypothetical resistance,

s

R '2 is split

into two components:

i. '2R , the per phase rotor resistance referred to stator itself, and

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ii.

s

sR

12' , the electrical equivalent of the mechanical load on the motor and is

generally called the load resistance RL or the dynamic resistance. The load resistance

RL depends upon the speed of the motor.

Figure 10 Modified Equivalent Circuit referred to stator representing rotor and the load resistances

The following approximate equivalent circuit is used as it simplifies the analysis of the induction

motor.

Figure 11 Approximate Equivalent Circuit of 3-phase induction motor

4. a. Write a brief note on double cage rotor induction motors. (8) (or) Explain the

advantages of double cage induction motor over single cage induction motor a double

cage induction motor. (8)

The main disadvantage of the squirrel cage induction motor is lower starting torque. But

the slip ring induction motor has the ability to increase the starting torque by inserting additional

resistance in to the rotor circuit. However such a procedure cannot be adopted for a squirrel cage

motor because its cage is permanently short-circuited. In order to provide high starting torque at

low starting current, double-cage construction is used.

Construction

Figure 12 Double Cage Induction Motor

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The outer winding consists of bars of smaller cross-section short-circuited by end rings.

Therefore, the resistance of this winding is high. Since the outer winding has relatively open

slots and a poorer flux path around its bars, it has a low inductance. Thus the resistance of the

outer squirrel-cage winding is high and its inductance is low.

The inner winding consists of bars of greater cross-section short-circuited by end rings.

Therefore, the resistance of this winding is low. Since the bars of the inner winding are

thoroughly buried in iron, it has a high inductance. Thus the resistance of the inner squirrel cage

winding is low and its inductance is high.

Operating Principle:

When a rotating magnetic field sweeps across the two windings, equal e.m.f.s are induced in

each.

At starting, the rotor frequency is the same as that of the line (i.e., 50 Hz), making the

reactance of the lower winding much higher than that of the upper winding. Because of the

high reactance of the lower winding, nearly all the rotor current flows in the high-resistance

outer cage winding. This provides the good starting characteristics of a high-resistance cage

winding. Thus the outer winding gives high starting torque at low starting current.

As the motor accelerates, the rotor frequency decreases, thereby lowering the reactance of the

inner winding, allowing it to carry a larger proportion of the total rotor current At the normal

operating speed of the motor, the rotor frequency is so low (2 to 3 Hz) that nearly all the rotor

current flows in the low-resistance inner cage winding. This results in good operating

efficiency and speed regulation.

The starting torque of this motor ranges from 200 to 250 percent of full-load torque with a

starting current of 4 to 6 times the full-load value. It is classed as a high-torque, low starting

current motor.

b. Write a brief note on induction generator. (8)

When an induction machine runs at a speed greater than its synchronous speed with the

help of prime mover, then it is called an induction generator. But it cannot function as an isolated

place without any supply, because it generates only when excitation comes from the stator side.

These generators are used in low capacity power stations, most frequently in micro hydro power

stations and wind power stations.

Figure 13 Self Excited Induction Generator

But, the external reactive source must remain permanently connected to the stator windings

responsible for the output voltage control. In interconnected applications, the synchronous grid

supplies such reactive power. In stand-alone applications, the reactive power must be supplied by

the load itself, or by a bank of capacitors connected across its terminals, or by an electronic

13


inverter. When capacitors are connected to induction generator, the system is usually called a

SEIG (a self-excited induction generator).

Advantages:

It does not hunt or drop out of synchronism.

Simple and rugged construction.

Cheaper in cost

Easy maintenance

Disadvantages:

Cannot be operated independently

Can deliver only leading current

May affect the system stability

5. a. Sketch and explain the torque slip characteristics of the 3-pahse cage and slip ring

induction motors. Show the stable region in the graph. (10)

Squirrel Cage Induction Motor:

When the load is applied to the shaft of the squirrel cage induction motor, the following

reactions will be observed:

Speed decreases

Slip increases

Rotor induced e.m.f increases consequently the rotor current increases

Torque developed by the rotor increases until it is sufficient to carry the applied load

The motor continues to run at that slip which will develop the required torque for that

particular load.

The motor may be loaded continuously till pull-out torque developed, at which point the

motor will stop if more load is placed on it. Pull-out occurs at a slip well beyond the normal

range operation of the squirrel cage induction motor. Even under normal operation, however, a

suddenly applied load may require the temporary development of this maximum torque, after

which the motor will speed up to its full load value. Thus, in a manner similar to the shunt motor,

the induction motor inherently adjusts itself to the applied external torque.

Slip ring Induction motor:

Figure 14 Slip-ring Induction Motor with Variable External Resistance at Rotor Circuit

Figure 14 shows the slip-ring induction motor with a variable resistance at its rotor. If the

resistance value is zero, then the motor will act same as squirrel cage induction motor. Under

normal running condition, if the rotor resistance is doubled, the following reactions will be

observed:

The rotor current will become half of its previous condition since the rotor speed, and

hence the rotor voltage, cannot change instantaneously.

The developed torque decreases due to reduced rotor current.

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The torque will increase even when there is no change in load, the motor slows down.

Slip increases

The rotor induced voltage increases

The rotor current now increases until it is sufficient to develop enough torque to again

carry the load at a constant speed.

The torque developed at standstill is higher with increased rotor resistance. This is

explained by the fact that at standstill the rotor frequency is equal to the line frequency

and rotor resistance is maximum. The addition of rotor resistance obviously improves the

powerfactor in the circuit while reducing the rotor current. Since standstill rotor

resistance is much greater than its resistance, the powerfactor increases more rapidly than

the decrease of rotor current, and hence the developed torque is greater with the added

resistance.

Figure 15 Torque Slip Characteristics of Induction Motors

b. Describe the principle of operation of synchronous induction motor. (6)

If the rotor of a slip-ring

induction motor is fed from a DC

power, this motor can be run at a

synchronous speed. Such motors are

called synchronous induction

motors.

Consider a normal slip-ring

induction motor with a conventional

three phase rotor winding is

energized by a fixed DC excitation.

Hence the pole axes of due to DC

excitation are also fixed and do not

change like the poles due to three

phase AC excitation. These fixed

rotor poles get magnetically locked

Figure 16 Connection of Synchronous Induction Motor

Stable Region Unstable Region

Slip Ring Im

Cage Im

15


with the rotating magnetic field of stator supplied from three phase AC supply. Therefore the

motor runs at a constant speed equal to synchronous speed.

Figure 15 shows the connection diagram of synchronous induction motor. The motor is

started as normal slip ring induction motor with additional resistance inserted at the rotor through

the slip ring when the switch is at the Start position. When the resistance is completely removed

from the rotor circuit, the switch is changed to Run position. Hence the exciter is connected to

the rotor winding, and the machine with DC excited rotor pulls into synchronism just like the

synchronous motor.

This motor can be stated with load as slip ring induction motor and can be operated at

constant speed and improved power factor as synchronous motor.

Advantages:

1. It starts and synchronizes itself against heavy loads

2. The exciter used can be small due to relatively smaller airgaps.

3. No separate damper winding is required.

Important Formulae

Synchronous Speed,

r.p.m p

fN s

120

Slip:

100

s

rs

N

NNSlip%

Nr-Rotor speed in r.p.m

Frequency of Rotor current, sff '

s-slip and f- Supply frequency

Full load Torque:

22

22

222

sXR

RksET f

Starting Torque:

22

22

222

XR

RkETst

Maximum Torque:

22

22

2X

kETm

16


Condition for maximum torque:

2

222

X

RsorXsR mm )(

Ratio of Full load torque to maximum torque:

22

2

m

m

m

f

ss

ss

T

T

Ratio of starting torque to maximum torque:

12

2

m

m

m

st

s

s

T

T

Rotor Copper loss from Rotor Gross output and vice-versa:

s

s

output gross Rotor

Loss Copper Rotor

1

Rotor input from Rotor output and vice-versa:

sInput Rotor

Output Rotor1

Rotor Copper loss from Rotor Input and vice-versa: Input RotorsLoss Copper Rotor

Equivalent Circuit Equations:

'2101 RRR

'2101 XXX

1

12

sRRL''

0101

2jXRR

VI

L

ph

'

'

m

ph

X

VI 0

'201 III

01

1Z

VI

ph

s

RIP

'' 222 3

''223 RILoss Copper Rotor

21 PsP Power, Mechanical m

s

sRIP Developed, Power Gross g

13 22

''

.

''''

NmN

sRI

N

s

sRI

T Torque, Grosss

g

60

2

13

60

2

13 2222

University Question Problems

1. An induction motor has an efficiency of 0.9 when the shaft load is 45kW. At this load, stator

ohmic loss and rotor ohmic loss each is equal to the iron loss. The mechanical loss is one-

third of the no-load losses. Neglect the ohmic losses at no-load. Calculate the slip. (8)

17


2. A 50 HP, 6–Pole, 50 Hz, slip ring IM runs at 960 rpm on full load with a rotor current of 40

A. Allow 300 W for copper loss in S.C. and 1200 W for mechanical losses, find R 2 per

phase of the 3- phase rotor. (6)

3. An 18.65 KW, 4 pole, 50 Hz, 3 phase induction motor has friction and windage losses of

2.5% of the output. Full load slip is 4%. Find for full load (1) rotor copper loss, (2) rotor

input, (3) shaft torque and (4) the gross electromagnetic torque. (8)

4. A 6 pole, 50 Hz, 3-Phase, induction motor running on full load develops a useful torque of

160 N-m. When the rotor emf makes 120 complete cycle per minute. Calculate the shaft

power input. If the mechanical torque lost in friction and that for core loss is 10 N m,

compute

(1) The copper loss in the rotor windings.

(2) The input the motor.

(3) The efficiency.

The total stator loss in given to be 800 W. (8)

5. A 746kW, 3-phase 50Hz, 16 Pole induction motor has a rotor impedance of (0.02+j0.15) Ω

at standstill. Full load torque is obtained at 360 r.p.m. Calculate,

(i) the ratio of maximum to full load torque

(ii) The speed at maximum torque and

(iii) The rotor resistance to be added to get maximum starting torque. (12)

6. A 15kW, 400V, 50Hz, 3-Phase star connected induction motor gave the following test

results:

No-load test: 400V, 9A, 1310W

Blocked Rotor Test: 200V, 50A, 7100W

Stator and rotor ohmic losses at standstill are assumed equal.

Draw the induction motor circle diagram and calculate,

(i) Line Current

(ii) Powerfactor

(iii) Slip

(iv) Torque and efficiency at full load. (16)

7. A 4 Pole 50Hz, 7.46kW motor has, at rated voltage and frequency, at starting Torque of

160% and maximum torque of 200% of full load torque. Determine,

(i) Full load torque

(ii) Speed at maximum torque (8)

8. A 3-phase, star connected 400V, 50Hz, 4 Pole induction motor has the following per phase

parameters in ohms, referred to the stator.

R1=0.15, X1=0.45, R2'=0.12, X2'=0.45, Xm=28.5

Compute the stator current and power factor when the motor is operated at rated voltage and

frequency with Slip,s=0.04. (8)

9. A 3-phase IM has starting torque 100% and a maximum torque of 200% of full load torque.

Find slip at maximum torque. (6)

10. A 100kW (output), 3300V, 3-phase, star connected induction motor has a synchronous speed

of 1500 r.p.m. The full load slip is 1.8% and full load powerfactor 0.85. Stator copper loss =

2440W. Iron loss=3500W. Rotational losses=1200W. Calculate (1) the rotor copper loss, (2)

the line current and (3) the full load efficiency. (8)

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11. Draw the circle diagram of a 15 HP, 230V, 50Hz, 3 - phase slip ring induction motor with a

star connected stator and rotor. The winding ratio is unity. The stator resistance is 0.42 Ohm

per phase and the rotor resistance is 0.3 Ohm per phase. The following are the test readings,

No-load test: 230V, 9A, p.f=0.2143

Blocked Rotor Test: 115V, 45A, p.f=0.454

Find,

a. Starting Torque

b. Maximum Torque

c. Maximum Powerfactor

d. Slip for Maximum Torque

e. Maximum Power Output. (16)

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Unit 3 - Induction Motor PART - A 1. 2. - Rajkumar Kuppapillai · PDF fileUnit 3 - Induction...

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Transcript of Unit 3 - Induction Motor PART - A 1. 2. - Rajkumar Kuppapillai · PDF fileUnit 3 - Induction...