THREE-PHASE

INDUCTION MOTOR

Part-2

Torque Under Running Conditions )(2cosor;2cos2 rErITrIET

where, E2= rotor emf/phase under standstill conditions

Ir= rotor current/phase under running conditions

2sErE Now,

2)2

(22

2sXR

sE

rZrE

rI

2)

2(2

2

22

cossXR

R

2)2

(22

222)

2(2

2

22)

2(2

2

2sXR

RsEk

sXR

R

sXR

sET

2)2

(22

2221sXR

RsEkT

)( rEAlso,

SN

k 23

1where,

The above torque expression can be written as follows:

2)2

(22

2where,2

222

23

2)2

(22

222

23 sXRrZ

rZ

RsE

SNsXR

RsE

SN

T

22

22

2221

1when,

XR

REkT

s

Example 34.10 The star connected rotor of an induction motor has a standstill impedance of (0.4+j4) ohm/phase and the rheostat impedance is (6+j2) ohm/phase. The motor has an induced emf of 80 V between slip-rings at standstill when connected to its normal supply voltage. Find:

(i) rotor current at standstill with the rheostat is in the circuit.

(ii) When the slip-rings are short-circuited and motor is running with a slip of 3%.

Solution: (1) Standstill conditions

Voltage/rotor phase = 80/3 =46.2 V

Rotor and starter impedance/phase= ((0.4+j4)+ (6+j2)= (6.4+j6)=8.7743.15o.

Rotor current/phase=46.2/8.77 5.27 A (power factor =cos43.15o =0.729)

(2) Running Conditions: Here starter impedance is cutout.

Rotor voltage/phase, Er=sE2= 0.0346.2= 1.386 V

Rotor reactance/phase, Xr=sX2=0.034=0.12 ohm

Rotor impedance/phase, Zr= 0.4+j0.12=0.417616.7o.

Rotor current/phase =1.386/0.4176=3.32 (power factor =cos16.7o =0.96)

Condition for Maximum Torque

Under Running Conditions The torque of a rotor under running conditions is 2)

2(2

2

222

12)2

(22

22sXR

RsEk

sXR

RsEkT

The condition for maximum torque may be obtained by differentiating the above expression with respect to slip s and then putting it equal to zero.

However, it is simpler to put Y=1/T and than differentiate it.

22

22

2

2

22

2)2

(

22

22

22

2)2

(22

REk

sX

sEk

R

RsEk

sX

RsEk

R

RsEk

sXRY

022

22

222

REk

X

Esk

R

dsdY

22;2

222

2;0

22

22

222 sXRXsR

REk

X

Esk

R

Hence, torque under running condition is maximum at that value of the slip s which makes rotor reactance per phase equal to rotor resistance per phase.

This slip is sometimes written as sb and the maximum torque Tb.

Slip corresponding to maximum torque is s=R2/ X2.

Putting R2= s X2 in the above equation for the torque, we get

)2

22or(

22

22)

2(2

22

22)

2(2)

2(

)2

(2

max R

sEk

X

Ek

sX

XsEk

sXsX

sXsEkT

Substituting value of s=R2/ X2 in the

torque equation, we get 2

2

221

22

2)2

/2

(22

222

)2

/2

(1

max X

Ek

XXRR

REXRkT

Since , we have N-m. S

Nk 2

31

22

22

23

max X

E

SN

T

From the above, it is found

1. That the maximum torque is independent of rotor resistance as such.

2. However, the speed or slip at which maximum torque occurs is determined by the rotor resistance. As seen from above, torque becomes maximum when rotor reactance equals its resistance. Hence, by varying rotor resistance (possible only with slip-ring motors) maximum torque can be made to occur at any desired slip (or motor speed).

3. maximum torque varies inversely as standstill reactance. Hence, it should be kept as small as possible.

4. maximum torque varies directly as the square of the applied voltage.

5. for obtaining maximum torque at starting (s=1), rotor resistance must be equal to rotor reactance.

Relation Between Torque and Slip A family of torque/slip curves is shown in Fig. 32.21 [1] for a range of s= 0 to

s=1 with R2 as the parameter. We have seen that 2)2(2

2

22

sXR

RsEkT

It is clear that when s=0, T=0, hence the curve starts from point 0. At normal speeds, close to synchronism, the term (sX2) is small and hence

negligible w.r.t. R2.

2R

sTor if R2 is constant.

Hence, for low values of slip, the torque/slip

curve is approximately straight line.

As slip increases (for increasing load on the motor), the torque also increases and becomes maximum when s=R2/X2.

This torque is known as ‘pull-out’ or ‘breakdown’ toque Tb or ‘stalling torque’.

1.0 0.8 0.6 0.4 0.2 0.00.00

0.02

0.04

0.0 0.2 0.4 0.6 0.8 1.0

Torq

ue [N

-m]

Slip

R2=2 ohm

R2=3 ohm

R2=4 ohm

K=0.3; E2=100 V; X

2=10 ohm; =0.03 Wb

Speed % of Ns

As the slip further increases (i.e. motor speed falls) with further increases in motor load, then R2 becomes

negligible as compared to (sX2).

Therefore, for large values of slip ssX

sT

12)2(

Hence, the torque/slip curve is a rectangular hyperbola.

So, we see that beyond the point of maximum torque, any further increase in motor load results in decrease of torque developed by the motor.

The result is that the motor slows down and eventually stops.

The circuit-breakers will be tripped open if the circuit has been so protected.

In fact, the stable operation of the motor lies between the values of s=0 and that corresponding to maximum torque.

It is seen that although maximum torque does not depend on R2, yet

the exact location of Tmax is dependent on it.

Greater the R2 greater is the value of slip at which the maximum

torque occurs.

1.0 0.8 0.6 0.4 0.2 0.00.00

0.02

0.04

0.0 0.2 0.4 0.6 0.8 1.0

Tor

que

[N-m

]

Slip

R2=2 ohm

R2=3 ohm

R2=4 ohm

K=0.3; E2=100 V; X

2=10 ohm; =0.03 Wb

Speed % of Ns

Effect of Change in Supply Voltage on

Torque and Speed We have seen that 2)2(2

2

22

sXR

RsEkT

As where V is the supply voltage. Hence 2sVTVE 2

Obviously, torque at any speed is proportional to the square of the applied voltage.

If stator voltage decreases by 10%, the torque decreases by 20%.

Changes in supply voltage not only affect the starting torque Tst but torque under

running conditions also.

If V decreases, then T also decreases. Hence, for maintaining the same torque, slip

increases i.e. speed falls.

Let V change to V’, s to s’ and T to T’; then

2''

2

' Vs

sV

T

T

Effect of Changes in Supply Frequency

on Torque and Speed Any important changes in frequency take place on a large distribution system except during a major distribution.

However, large frequency changes often take place on isolated, low-power systems in which electric energy is generated by means of diesel engines or gas turbines.

Example of such systems are: emergency supply in a hospital and the electric system on a ship etc.

The major effect of change in supply frequency is on motor speed.

If frequency drops by 10%, then motor speed also drops by 10%.

Machine tools and other motor-driven equipment meant for 50Hz causes problem when connected to 60Hz supply.

Everything runs (60-50)100/50=20% faster than normal and this may not be acceptable in all applications. In that case, we have to use either gears to reduce motor speed or an expensive 50 Hz source.

A 50 Hz motor operates well on a 60 Hz line provided its terminal voltage is raised to 60/50=6/5 (i.e. 120%) of the name-plate rating.

In that case, the new breakdown torque becomes equal to the original breakdown torque and the starting torque is only slightly reduced.

However, power factor, efficiency and temperature rise remain satisfactory.

Similarly, a 60 Hz motor can operate satisfactory on 50 Hz supply provided its terminal voltage is reduced to 5/6 (i.e. 80%) of its name-plate rating.

Full-Load Torque and Maximum Torque Let sf be the slip corresponding to full-load torque, then

2)2(22

2

XfsR

RfsfT

22

1maxand

XT

2)2(22

222

max XfsR

XRfs

TfT

Dividing both the numerator and the

denominator by X22, we get 22)2/2(

2/22

maxfsXR

XRfs

TfT

Let a=R2/X2 =resistance/reactance, then 22

2

maxfsa

afs

TfT

222

torquemaximum

slipanyattorqueoperatinggeneral,In

sa

sas

Starting Torque and Maximum Torque

22

22

2

XR

RstT

22

1maxand

XT

21

22)2/2(1

2/22

max a

a

XR

XR

TstT

The staring torque Tst and the maximum torque Tmax can be written as follows:

phaseperreactancerotor

erestistancrotor

2

2where, X

Ra

Variations in Rotor Resistance

The magnitude of the rotor current varies with load carried by the motor.

sN

NinputRotor

outputRotorthat,seenweAs

sN

N inputRotoroutputRotoror,

outputRotorinputRotor N

sN KNTTN 2outputRotorAlso,

TsKNKNTN

sN inputRotor s

inputRotor

lossCuRotorNow,

s

RI 2223

inputRotor

TsKNs

RI 2

223

s

RIT

222 1since2

22 sRstIstT

sliploadFull2

22 fs

fs

RfIfT

22

fIstI

fsfTstT

where, I2st and I2f are the rotor currents for starting and full-load running conditions.

Induction Motor as a Generalized Transformer The transfer of energy from stator to the rotor of an induction motor takes place entirely inductively, with the help of flux mutually linking the two.

Hence, an induction motor is essentially a transformer with stator forming the primary and rotor forming (the short-circuited) rotating secondary (Fig. 32.45).

The vector diagram is similar to that of a

transformer. In the vector diagram of Fig. 32.46 V1 is the applied

voltage per stator phase.

R1 and X1 are stator resistance and leakage reactance

per phase respectively, shown external to the stator winding in Fig. 32.45.

The applied voltage V1 produces a magnitude flux

which links both primary and secondary thereby producing a counter emf of self-induction E1 in primary

(i.e. stator) and a mutually induced emf Er (sE2) in

secondary (i.e. rotor). There is no secondary voltage V2 in secondary because

whole of the induced emf Er is used up in circulating

the rotor current as the rotor is closed upon itself (which is equivalent to its being short-circuited).

111111Obviously, XjR IIEV The magnitude of Er depends on voltage transformation ratio K between stator

and rotor and the slip. As it is wholly absorbed in the rotor impedance. 222222 XjRr IIZIE

In the vector diagram I0 is the no-load primary current.

It has two components (i) the working or iron loss components Iw which supplies the no-load motor losses

and (ii) the magnetizing component I which sets up

magnetic flux in the core and the air gap. 22

0Obviously, IwII

In Fig. 32-45, Iw and I are taken care of by an exciting

circuit containing R0=E1/Iw; and X0=E1/I , respectively.

Even though the frequencies of stator and rotor currents are different, yet magnetic fields due to them are synchronous with each other, when seen by an observer stationed in space- both field rotate at synchronous speed Ns. The current following in the short-circuited rotor produces a magnetic field, which revolves round the rotor in the same direction as the stator field. The speed of rotation of the rotor field is

NsNsN

NsNsNssN

P

sf

Prf

rN

120120

sNsrNsNN )1(,speedRotor Hence, speed of the rotating field of the rotor with respect to the stationary stator or space is sNNNsNNssN )(

Rotor Output Primary current I1 consists of two parts, I0 and I2’. It is the later which is transferred

to the rotor.

If the applied primary voltage V1, some is absorbed in the primary itself (=I1Z1) and

the remaining E1 is transferred to the rotor.

If the angle between E1 and I2’ is , then cos'

21einput/phasRotor IE

cos'213inputrotorTotal IE

The electrical input to the rotor which is wasted in the form of heat is

)2223orcos23 RIrEI

s 1inputrotor

outputrotor

inputrotorlossCurotor s

SpeedsSynchronou

SpeedActual1efficiencyrotor

sN

Ns

The induction motor is called asynchronous motor because the rotor of an induction motor does not run at synchronous speed (the speed of rotation of the magnetic flux, produced by the primary winding of a dynamo).

Equivalent Circuit of the Rotor When motor is loaded, the

rotor current I2 is given by )1(

2)2(22

2

sXR

EsrI

)2(

2)2(2)/2(

2

XsR

ErI

The rotor circuit which actually consists of a fixed resistance R2 and a variable

reactance sX2 (proportional to slip) connected across Er=sE2 [Fig.32.47(a)and Eq

(1)] can be looked upon as equivalent to a rotor circuit having a fixed reactance X2

connected in series with a variable resistance R2/s (inversely proportional to slip)

and supplied with constant voltage E2 [Fig. 32.47(b) and Eq. (2) ].

)3(11

222

sRR

s

RIt can be written as:

It consists of two parts:

1. The first part R2 is the rotor resistance itself and represents the rotor Cu

loss.

1

12 s

R2. The second part is .

This is known as the load resistance RL and is the electrical equivalent of the

mechanical load on the motor.

In other words, the mechanical load on an induction motor can be represented

by a non-inductive resistance of the value .

1

12 s

R

The equivalent rotor circuit along with the load resistance RL may be drawn as in Fig. 32.48.

Equivalent Circuit of an Induction Motor As in the case of a transformer, in this case also, the secondary values may be transferred to the primary and vice versa.

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.

The equivalent circuit of an induction motor where all values have been referred

to primary i.e. stator is shown in Fig. 32. 49.

As shown in Fig. 32.50, the exciting circuit may be transferred to the left, because inaccuracy involved is negligible but the circuit and hence the calculations are very much simplified.

This is known as the approximate

equivalent circuit of the induction motor.

If transformation ratio is assumed unity i.e. E2/E1=1, then the equivalent

circuit is as shown in Fig. 32.51

instead of that in Fig. 32.49b.

Power Balance Equation With respect to Fig. 32.49(a), following power

relations in an induction motor can be deduced:

1cos113powerInput IV 121lossCuStator RI

02lossCoreStator RwI

sRI /'2

2'23rotortodtransferrePower '

22'23lossCuRotor RI

Mechanical power developed by rotor (Pm) or gross power developed by rotor (Pg)

watt1'

22'23'

22'23/'

22'23lossesCuRotorinputRotor

s

sRIRIsRI

If Tg is the gross torque developed by the rotor, then

watt1'

22'23'

22'23/'

22'23

60

2

s

sRIRIsRI

NgTgT

m-N2

601'2

2'23

Ns

sRIgT

)1(Now, ssNN

Hence gross torque becomes m-N/'

22'23

2

60

)1(2

601'2

2'23

sN

sRI

ssNs

sRIgT

m-N'2

2'23

55.9/'

22'23

55.9s

R

sN

I

sN

sRIgT

Since gross torque in synchronous watts is equal to

the power transferred to the rotor across the air-gap. wattSynch./'2

2'23 sRIgT

It is seen from the approximate

circuit of Fig. 32.50 that )'21()/'

21(

1'2

XXjsRR

VI

m-N'2

2)'21(2)/'

21(

213

55.9s

R

XXsRR

V

sNgT

The gross torque Tg is different from shaft torque, which is

less than Tg by the torque required to meet windage and

frictional losses.

Maximum Power Output Fig. 32.52 shows the approximate equivalent circuit of an induction motor with the simplification that (i) exciting circuit is omitted i.e. I0 is neglected and (ii) K is assumed unity.

As seen, gross power output for 3-phase induction motor is

LRIgP 213

201

2)01(

11Now,

XLRR

VI

201

2)01(

213

XLRR

LRVgP

The condition for maximum power output can be found be differentiating the above equation with respect to load resistance RL and by equating the first

derivative to zero. If it is done, it will be found that

0201

2)01(

XLRR

LR

LdRgPd

0201

2)01(

213

XLRR

LR

LdRgPd

VLdRgPd

02]2

012)01[(

]201

2)01[(]2

012)01[(

XLRR

LdR

XLRRdLR

LdRLdR

XLRR0

201

2)01(

XLRR

LR

LdR

gPd

0201

2)01[(]2

012)01[(

LdR

XLRRdLR

LdRLdR

XLRR

0]201

2)01([]2

012)01[(

LdR

dX

LdRLRRd

LRXLRR

0])01(2[]201

2)01[( LdRLdR

LRRLRXLRR

0)01(2]201

2)01[( LRRLRXLRR

022012201

20122

01 LRRLRXLRLRRR

02201

201 LRXR 2

01201

201

2 ZXRLR

01ZLR

where, Z01 = leakage

impedance of the motor as

referred to primary

Hence, the power output is maximum when the equivalent load resistance is equal to the standstill leakage impedance of the motor.

Corresponding Slip

1

12that,knowWe

sRLR

01power,outputmaximumFor ZLR

1

1201 s

RZ201

2Hence,RZ

Rs

This is the slip corresponding to

maximum gross power output.

201

2)01(

213

that,knowWeXLRR

LRVgP

The value of Pgmax is obtained by

substituting RL by Z01 in the above

equation.

201201012

012

13max

ZZR

ZVgP

201

2)0101(

012

13max

XZR

ZVgP

201

201

201010122

01

012

13max

RZZZRR

ZVgP

)0101(2

213

012012

213

max ZR

V

ZR

VgP

It should be noted that V1 is voltage/phase of the motor and K

has been taken as unity.

Example 34.35 A 100 kW (output), 3300 V, 50 Hz, 3-phase, star connected induction motor has a synchronous speed of 500 rpm. The full load slip is 1.8% and full-load power factor 0.85. Stator copper loss= 2440 W. Iron loss= 3500 W. Rotational loss= 1200 W. Calculate: (i) The rotor copper loss, (ii) the line current, and (iii) the full-load efficiency.

Solution: Pm= output + rotational losses = 100+1.2=101.2 W

kW855.12.101018.01

018.01

LossCuRotor(i)

mPs

s

(ii) Rotor input, R2 = Pm + rotor Cu loss

= 101.2+1.855=103.055 kW

Stator input= P2 + stator Cu and iron losses

=103.055+2.44+3.5=108.995 kW

(iii) Full-load efficiency= 100,000/108,995=0.917 or 91.7%

Example 34.36 The power input to the rotor of a 440 V, 50 Hz, 6 pole, 3-phase induction motor is 100 kW. The rotor electromotive force is observed to make 120 cycles per minute. Calculate: (i) the slip, (ii) the rotor speed, (iii) mechanical power developed, (iv) the rotor Cu loss per phase, and (v) speed of a stator field with respect to rotor.

Solution: (i) we know that, 04.050/)60/120('/So, ffssff '

(ii) Ns= 120f/P=12050/6=1000 rpm; and

N= (1-0.04) 1000=960 rpm

(iii) Pm= (1-s)P2=(1-0.04) 100=96 kW

(iv) Total Cu loss= sP2=0.04100 = 4 kW;

Cu loss/phase= 4/3 = 1.333 kW

(v) Speed of the stator field with respect to rotor is = 1000 – 960

= 40 rpm

Example 34.38 A 400 V, 50 Hz, 6-pole, -connected, 3- induction motor consumes 45 kW with a line current of 75 A and runs at a slip of 3%. If stator iron loss is 1200 W, windage and friction loss is 900 W and resistance between two terminals is 0.12 . Calculate: (i) power supplied to the rotor, P2, (ii) rotor Cu loss, Pcr, (iii) power supplied to the load, Pout, (iv) efficiency, and (v) shaft torque developed.

Solution: lag866.0754003

100045cos

A line current of 75A means a phase-current of 75/3 i.e. 43.3A

Next, winding resistance has to be worked out Refer to Fig. 34.40.

r and 2r are in parallel have an equivalent resistance measured at a and b terminals in delta connected motor as (r×2r/3r)=2r/3 ohms

From the data given, (2r/3)=0.12, r=0.18 ohm

Total stator Cu loss= 3×43.32×0.18=1012 W

Total input to stator= 45,000 W

Stator output= 45,000-(1012+1200)=42,788 W

Rotor Cu Loss= Slip × Rotor input= 0.03 × 42,788=1248 W

Rotor mechanical output power= 42,788-1248= 41,504 W

Shaft output= Mechanical output of rotor – Mechanical Loss

=41,504-900= 40,604 W

%23.92%100000,45604,40Efficiency

mN4009702

60604,40 rque,output toShaft

T

Example 34.41 An 18.65 kW, 4 pole, 50 Hz, 3 phase induction motor has friction and windage losses of 2.5% of the output. The full-load slip is 4%. Compute for full-load: (i) the rotor Cu loss, (ii) the rotor input, (iii) the shaft torque, and (iv) the gross electromagnetic torque.

Solution: Motor output, Pout= 18,650 W

Friction and windage loss, Pw=2.5% of 18,650 = 466 W

Rotor gross output, Pm= 18,650+466 =19,116 W

W6.796116,1904.01

04.0mpower,output gross

1lossCuRotor

;1mpower,output gross

lossCuRotor)(

Ps

ss

sP

i

5.912,1904.0

6.796lossCuRotor2

input,Rotor;2

input,RotorlossCuRotor)( sPs

Pii

(iii) Tsh=9.55Pout/N= 9.5518,650/1440=123.7 N-m

(iv) Gross torque, Tg=9.55Pm/N= 9.5519,116/1440=126.8 N-m

or Tg=P2/2Ns=19,91318,650/(225)=127 N-m

No-Load TestWithout connecting any load on the induction motor shift, full voltage is applied across the stator windings terminals. Since output of the motor at no-load is zero, the whole of the input power is wasted as various losses.At no-load, the speed of the rotor is very nearly equal to the synchronous speed. The slip at no-load is very small and therefore the quantity RL=R2(1-s)/s, of the equivalent mechanical load is very high.The rotor induced emf and the rotor current, therefore, are negligibly small. The rotor for practical purpose may be considered as an open circuit.No-load test of an induction motor is, therefore, similar to no-load (or open-circuit) test on a transformer.

The losses at no-load are: (i) Cu-loss in the stator winding; (ii) Core (or iron) losses in the stator and rotor; and (iii) Frictional and windage losses.

From the total input at no-load, the Cu loss in the stator winding can be subtracted to get core loss plus friction losses.

The core-loss, and friction and windage losses at no-load are nearly the same as would occur under full load condition. So these losses are called fixed losses. This is because core loss depends on applied voltage, whereas friction and windage losses depend upon speed of the rotation of rotor.

Applied voltage is assumed to be constant and the variation of speed of an induction motor from no-load to full-load is negligibly small.

Connection diagram for no-load test on an induction motor is shown in Fig. 4.30.In practice, it is neither necessary nor feasible to run the induction motor synchronously so that the no-load test is carried out with different values of applied voltage, below and above the value of normal rated voltage.

The power input (Wo) is measured by two wattmeters, no-load current (Io) by an ammeter and applied voltage (Vo) by a voltmeter.

Fig. 4.30