1

Introduction to 3 phase induction motors

:Definition

Induction motor is an a.c. motor in which currents in the

stator winding (which is connected to the supply ) set up a flux which

causes currents to be induced in the rotor winding ; these currents

interact with the flux to produce rotation . Also called asynchronous

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

the secondary of a two winding transformer receiving its power from

primary. That is why an induction motor can be called as a rotating

transformer i.e., in which primary winding is stationary but the

secondary is free to rotate.

Types of induction motors: Depending on rotor there are two types of induction motors :-

:Squirrel cage induction motor -1 Rotors is very simple and consist of bars of aluminum (or copper)

with shorting rings at the ends.

wound rotor induction motor : -2

Three phase windings (star connected) with terminals brought out to

slip rings for external connections.

The first type is more common used compared to second one due to:

1- Robust : No brushes . No contacts on rotor shaft.

2- Easy to manufacture.

3- Almost maintenance-free , except for bearing and other

mechanical parts.

4- Since the rotor has very low resistance, the copper loss is low and

efficiency is high .

Construction:

There are two main types of components which are used in induction

motor manufacturing as follows:

- Active components : which are classified into two categories:-1

a- Magnetic materials (0.5 mm electrical steel)

b- Electrical materials ( copper wires ,insulations ,bars , end

rings ,slip rings ,brushes , and lead wires)

2

2- Constructional components: like frame , end shields , shaft ,

bearings , and fan . These components are shown in Figures 1 & 2 .

Figure- 1 parts of squirrel cage induction motor

Figure- 2 Axial view of squirrel cage induction motor

3

Stator construction :

The stator is made up of several thin laminations (0.5 mm )of

electrical steel (silicon steel) , they are punched and clamped

together to form a hollow cylinder ( stator core ) with slots , as

shown in Figure- 3. 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).The number

of poles of an induction motor depends on the internal connection

of the stator windings .

Rotor construction:

The squirrel cage rotor is made up of several thin electrical steel

lamination (0.5mm) with evenly spaced bars , which are made up

of aluminum or copper , along the periphery .In the most popular

type of rotor (squirrel cage rotor) , these bars are connected at ends

mechanically and electrically by the use of end rings as shown in

Fig.4. Almost 90 % of induction motors have squirrel cage rotors .

This is because the squirrel cage rotors has a simple and rugged

construction .The rotor slots are not exactly parallel to the shaft.

Instead , they are given a skew for two main reasons :The first

reason is to make the motor run quietly by reducing magnetic hum

and to decrease slot harmonics. The second reason is to help reduce

the locking tendency of the rotor (the rotor teeth tend to remain

locked under the stator teeth due to direct magnetic attraction

between the two). The rotor is mounted on the shaft using bearings

on each end ; one end of the shaft is normally kept longer than the

other for driving the load . Between the stator and the rotor, there

exist an air gap , through which due to induction , the energy is

transferred from the stator to the rotor.

4

The wound rotor has a set of windings on the rotor slots

which are not short circuited , but are terminated to a set of

slip rings . These are helpful in adding external resistors and

contactors , as shown in Figure 5.

5

Typical name plate of induction motor A typical name plate of induction motor is shown in Figure 6, and

Table 1 .

Motor standards NEMA : National Electrical Manufactures Association

IEC : International Electrotechnical Commission

6

Motor insulation class Insulations have been standardized and graded by their resistance to

thermal aging and failure. Four insulation classes are in common use ,

they have been designated by the letters A , B , F, and H .

The temperature capabilities of these classes are separated from each other

by 25 °C increments. The temperature capabilities of each insulation class

is defined as being the maximum temperature at which the insulation can

be operated to yield an average life of 20,000 hours , as in Table below.

Insulation Class Temperature Rating

A 105° C

B 130° C

F 155° C

H 180° C

Motor degree of protection

I P : International Protection , I P * #

Protection against ingress

of water # Protection against ingress

of bodies *

Non protected 0 Non protected 0

Protected against ingress of

dripping water .

1 Protected against ingress of

foreign solid bodies of 50

mm or greater.

1

Protection against ingress of

dripping water at maximum

angle of 150 degrees from

the vertical.

2 Protected against ingress of

foreign solid bodies of 12

mm or greater.

2

Protection against water

falling like rain.

3 Protected against ingress of

foreign solid bodies of 2.5

mm or greater.

3

Protection against splashing

water.

4 Protected against ingress of

foreign solid bodies of 1 mm

or greater.

4

Protection against water jets

.

5 Partially protected against

ingress of dust .

5

Protection against special

conditions on ship's board.

6 Totally protected against

ingress of dust.

6

Protection against

immersion in water .

7

Protection against

prolonged immersion in

water .

8

181

Chapter (8)

Three Phase Induction Motors

IntroductionThe three-phase induction motors are the most widely used electric motors inindustry. They run at essentially constant speed from no-load to full-load.However, the speed is frequency dependent and consequently these motors arenot easily adapted to speed control. We usually prefer d.c. motors when largespeed variations are required. Nevertheless, the 3-phase induction motors aresimple, rugged, low-priced, easy to maintain and can be manufactured withcharacteristics to suit most industrial requirements. In this chapter, we shallfocus our attention on the general principles of 3-phase induction motors.

8.1 Three-Phase Induction MotorLike any electric motor, a 3-phase induction motor has a stator and a rotor. Thestator carries a 3-phase winding (called stator winding) while the rotor carries ashort-circuited winding (called rotor winding). Only the stator winding is fedfrom 3-phase supply. The rotor winding derives its voltage and power from theexternally energized stator winding through electromagnetic induction andhence the name. The induction motor may be considered to be a transformerwith a rotating secondary and it can, therefore, be described as a “transformer-type” a.c. machine in which electrical energy is converted into mechanicalenergy.

Advantages(i) It has simple and rugged construction.(ii) It is relatively cheap.(iii) It requires little maintenance.(iv) It has high efficiency and reasonably good power factor.(v) It has self starting torque.

Disadvantages(i) It is essentially a constant speed motor and its speed cannot be changed

easily.(ii) Its starting torque is inferior to d.c. shunt motor.

182

Fig.(8.1)

8.2 ConstructionA 3-phase induction motor has two main parts (i) stator and (ii) rotor. The rotoris separated from the stator by a small air-gap which ranges from 0.4 mm to 4mm, depending on the power of the motor.

1. StatorIt consists of a steel frame which encloses ahollow, cylindrical core made up of thinlaminations of silicon steel to reducehysteresis and eddy current losses. A numberof evenly spaced slots are provided on theinner periphery of the laminations [See Fig.(8.1)]. The insulated connected to form abalanced 3-phase star or delta connectedcircuit. The 3-phase stator winding is wound for a definite number of poles asper requirement of speed. Greater the number of poles, lesser is the speed of themotor and vice-versa. When 3-phase supply is given to the stator winding, arotating magnetic field (See Sec. 8.3) of constant magnitude is produced. Thisrotating field induces currents in the rotor by electromagnetic induction.

2. RotorThe rotor, mounted on a shaft, is a hollow laminated core having slots on itsouter periphery. The winding placed in these slots (called rotor winding) may beone of the following two types:

(i) Squirrel cage type (ii) Wound type

(i) Squirrel cage rotor. It consists of a laminated cylindrical core havingparallel slots on its outer periphery. One copper or aluminum bar is placedin each slot. All these bars are joined at each end by metal rings called endrings [See Fig. (8.2)]. This forms a permanently short-circuited windingwhich is indestructible. The entire construction (bars and end rings)resembles a squirrel cage and hence the name. The rotor is not connectedelectrically to the supply but has current induced in it by transformer actionfrom the stator.

Those induction motors which employ squirrel cage rotor are calledsquirrel cage induction motors. Most of 3-phase induction motors usesquirrel cage rotor as it has a remarkably simple and robust constructionenabling it to operate in the most adverse circumstances. However, itsuffers from the disadvantage of a low starting torque. It is because therotor bars are permanently short-circuited and it is not possible to add anyexternal resistance to the rotor circuit to have a large starting torque.

183

Fig.(8.2) Fig.(8.3)

(ii) Wound rotor. It consists of a laminated cylindrical core and carries a 3-phase winding, similar to the one on the stator [See Fig. (8.3)]. The rotorwinding is uniformly distributed in the slots and is usually star-connected.The open ends of the rotor winding are brought out and joined to threeinsulated slip rings mounted on the rotor shaft with one brush resting oneach slip ring. The three brushes are connected to a 3-phase star-connectedrheostat as shown in Fig. (8.4). At starting, the external resistances areincluded in the rotor circuit to give a large starting torque. Theseresistances are gradually reduced to zero as the motor runs up to speed.

Fig.(8.4)

The external resistances are used during starting period only. When the motorattains normal speed, the three brushes are short-circuited so that the woundrotor runs like a squirrel cage rotor.

8.3 Rotating Magnetic Field Due to 3-Phase CurrentsWhen a 3-phase winding is energized from a 3-phase supply, a rotatingmagnetic field is produced. This field is such that its poles do no remain in afixed position on the stator but go on shifting their positions around the stator.For this reason, it is called a rotating Held. It can be shown that magnitude ofthis rotating field is constant and is equal to 1.5 φm where φm is the maximumflux due to any phase.

184

Fig.(8.5)

To see how rotating field is produced, consider a 2-pole,3i-phase winding as shown in Fig. (8.6 (i)). The threephases X, Y and Z are energized from a 3-phase sourceand currents in these phases are indicated as Ix, Iy and Iz[See Fig. (8.6 (ii))]. Referring to Fig. (8.6 (ii)), the fluxesproduced by these currents are given by:

)240t(sin

)120t(sintsin

mz

my

mx

°−ωφ=φ

°−ωφ=φωφ=φ

Here φm is the maximum flux due to any phase. Fig. (8.5) shows the phasordiagram of the three fluxes. We shall now prove that this 3-phase supplyproduces a rotating field of constant magnitude equal to 1.5 φm.

Fig.(8.6)

185

Fig.(8.7)

Fig.(8.8)

(i) At instant 1 [See Fig. (8.6 (ii)) and Fig. (8.6 (iii))], the current in phase X iszero and currents in phases Y and Z are equaland opposite. The currents are flowing outwardin the top conductors and inward in the bottomconductors. This establishes a resultant fluxtowards right. The magnitude of the resultantflux is constant and is equal to 1.5 φm asproved under:

At instant 1, ωt = 0°. Therefore, the threefluxes are given by;

( )

( ) mmz

mmyx

23240sin

;23120sin;0

φ=°−φ=φ

φ−=°−φ=φ=φ

The phasor sum of − φy and φz is the resultant flux φr [See Fig. (8.7)]. It isclear that:

Resultant flux, mmmr 5.123

2322

60cos232 φ=×φ×=°φ×=φ

(ii) At instant 2, the current is maximum(negative) in φy phase Y and 0.5 maximum(positive) in phases X and Y. Themagnitude of resultant flux is 1.5 φm asproved under:

At instant 2, ωt = 30°. Therefore, the threefluxes are given by;

2)210(sin

)90(sin230sin

mmz

mmy

mmx

φ=°−φ=φ

φ−=°−φ=φ

φ=°φ=φ

The phasor sum of φx, − φy and φz is the resultant flux φr

Phasor sum of φx and φz, 22120cos22' mm

rφ

=°φ×=φ

Phasor sum of φ'r and − φy, mmm

r 5.12 φ=φ+φ

=φ

Note that resultant flux is displaced 30° clockwise from position 1.

186

Fig.(8.9)

Fig.(7.10)

(iii) At instant 3, current in phase Z is zero andthe currents in phases X and Y are equal andopposite (currents in phases X and Y arc0.866 × max. value). The magnitude ofresultant flux is 1.5 φm as proved under:

At instant 3, ωt = 60°. Therefore, the threefluxes are given by;

( )( ) 0180sin

;2360sin

;2360sin

mz

mmy

mmx

=°−φ=φ

φ−=°−φ=φ

φ=°φ=φ

The resultant flux φr is the phasor sum of φx and − φy ( )0z =φQ .

mmr 5.1260cos2

32 φ=°φ×=φ

Note that resultant flux is displaced 60° clockwise from position 1.

(iv) At instant 4, the current in phase X ismaximum (positive) and the currents in phasesV and Z are equal and negative (currents inphases V and Z are 0.5 × max. value). Thisestablishes a resultant flux downward asshown under:

At instant 4, ωt = 90°. Therefore, the threefluxes are given by;

2)150(sin

2)30(sin

90sin

mmz

mmy

mmx

φ−=°−φ=φ

φ−=°−φ=φ

φ=°φ=φ

The phasor sum of φx, − φy and − φz is the resultant flux φr

Phasor sum of − φz and − φy, 22120cos22' mm

rφ

=°φ×=φ

Phasor sum of φ'r and φx, mmm

r 5.12 φ=φ+φ

=φ

Note that the resultant flux is downward i.e., it is displaced 90° clockwisefrom position 1.

187

It follows from the above discussion that a 3-phase supply produces a rotatingfield of constant value (= 1.5 φm, where φm is the maximum flux due to anyphase).

Speed of rotating magnetic fieldThe speed at which the rotating magnetic field revolves is called thesynchronous speed (Ns). Referring to Fig. (8.6 (ii)), the time instant 4 representsthe completion of one-quarter cycle of alternating current Ix from the timeinstant 1. During this one quarter cycle, the field has rotated through 90°. At atime instant represented by 13 or one complete cycle of current Ix from theorigin, the field has completed one revolution. Therefore, for a 2-pole statorwinding, the field makes one revolution in one cycle of current. In a 4-polestator winding, it can be shown that the rotating field makes one revolution intwo cycles of current. In general, fur P poles, the rotating field makes onerevolution in P/2 cycles of current.

∴ Cycles of current = 2P × revolutions of field

or Cycles of current per second = 2P × revolutions of field per second

Since revolutions per second is equal to the revolutions per minute (Ns) dividedby 60 and the number of cycles per second is the frequency f,

120PN

60N

2Pf ss =×=∴

or Pf120Ns =

The speed of the rotating magnetic field is the same as the speed of thealternator that is supplying power to the motor if the two have the same numberof poles. Hence the magnetic flux is said to rotate at synchronous speed.

Direction of rotating magnetic fieldThe phase sequence of the three-phase voltage applied to the stator winding inFig. (8.6 (ii)) is X-Y-Z. If this sequence is changed to X-Z-Y, it is observed thatdirection of rotation of the field is reversed i.e., the field rotatescounterclockwise rather than clockwise. However, the number of poles and thespeed at which the magnetic field rotates remain unchanged. Thus it is necessaryonly to change the phase sequence in order to change the direction of rotation ofthe magnetic field. For a three-phase supply, this can be done by interchangingany two of the three lines. As we shall see, the rotor in a 3-phase inductionmotor runs in the same direction as the rotating magnetic field. Therefore, the

188

Fig.(8.11)

Fig.(8.12)

direction of rotation of a 3-phase induction motor can be reversed byinterchanging any two of the three motor supply lines.

8.4 Alternate Mathematical Analysis for RotatingMagnetic Field

We shall now use another useful method to findthe magnitude and speed of the resultant flux dueto three-phase currents. The three-phase sinusoidalcurrents produce fluxes φ1, φ2 and φ3 which varysinusoidally. The resultant flux at any instant willbe the vector sum of all the three at that instant.The fluxes are represented by three variablemagnitude vectors [See Fig. (8.11)]. In Fig. (8.11),the individual flux directions arc fixed but their magnitudes vary sinusoidally asdoes the current that produces them. To find the magnitude of the resultant flux,resolve each flux into horizontal and vertical components and then find theirvector sum.

tsin2360cos)240tcos(60cos)120tcos(0

tcos23

60cos)240tcos(60cos)120tcos(ttcos

mmmv

m

mmmh

ωφ=°°−ωφ+°°−ωφ−=φ

ωφ=

°°−ωφ−°°−ωφ−ωφ=φ

The resultant flux is given by;

[ ] Constant5.123)tsin(tcos2

3mm

2/122m

2v

2hr =φ=φ=ω−+ωφ=φ+φ=φ

Thus the resultant flux has constant magnitude (= 1.5 φm) and does not changewith time. The angular displacement of φr relative tothe OX axis is

t

ttantcos2

3

tsin23

tanm

m

h

v

ω=θ∴

ω=ωφ

ωφ=

φφ

=θ

Thus the resultant magnetic field rotates at constant angular velocity ω( = 2 πf)rad/sec. For a P-pole machine, the rotation speed (ωm) is

sec/radP2

m ω=ω

189

Fig.(1-)

or .m.p.rinisN...f2P2

60N2

ss π×=

π

Pf120Ns =∴

Thus the resultant flux due to three-phase currents is of constant value (= 1.5 φm

where φm is the maximum flux in any phase) and this flux rotates around thestator winding at a synchronous speed of 120 f/P r.p.m.

For example, for a 6-pole, 50 Hz, 3-phase induction motor, N, = 120 × 50/6 =1000 r.p.m. It means that flux rotates around the stator at a speed of 1000 r.p.m.

8.5 Principle of OperationConsider a portion of 3-phase induction motor as shown in Fig. (8.13). Theoperation of the motor can be explained as under:

(i) When 3-phase statorwinding is energized froma 3-phase supply, arotating magnetic field isset up which rotates roundthe stator at synchronousspeed Ns (= 120 f/P).

(ii) The rotating field passesthrough the air gap andcuts the rotor conductors,which as yet, arestationary. Due to the relative speed between the rotating flux and thestationary rotor, e.m.f.s are induced in the rotor conductors. Since therotor circuit is short-circuited, currents start flowing in the rotorconductors.

(iii) The current-carrying rotor conductors are placed in the magnetic fieldproduced by the stator. Consequently, mechanical force acts on the rotorconductors. The sum of the mechanical forces on all the rotor conductorsproduces a torque which tends to move the rotor in the same direction asthe rotating field.

(iv) The fact that rotor is urged to follow the stator field (i.e., rotor moves inthe direction of stator field) can be explained by Lenz’s law. Accordingto this law, the direction of rotor currents will be such that they tend tooppose the cause producing them. Now, the cause producing the rotorcurrents is the relative speed between the rotating field and the stationaryrotor conductors. Hence to reduce this relative speed, the rotor startsrunning in the same direction as that of stator field and tries to catch it.

190

8.6 SlipWe have seen above that rotor rapidly accelerates in the direction of rotatingfield. In practice, the rotor can never reach the speed of stator flux. If it did,there would be no relative speed between the stator field and rotor conductors,no induced rotor currents and, therefore, no torque to drive the rotor. Thefriction and windage would immediately cause the rotor to slow down. Hence,the rotor speed (N) is always less than the suitor field speed (Ns). This differencein speed depends upon load on the motor.

The difference between the synchronous speed Ns of the rotating stator field andthe actual rotor speed N is called slip. It is usually expressed as a percentage ofsynchronous speed i.e.,

% age slip, 100NNNs

s

s ×−

=

(i) The quantity Ns − N is sometimes called slip speed.(ii) When the rotor is stationary (i.e., N = 0), slip, s = 1 or 100 %.(iii) In an induction motor, the change in slip from no-load to full-load is

hardly 0.1% to 3% so that it is essentially a constant-speed motor.

8.7 Rotor Current FrequencyThe frequency of a voltage or current induced due to the relative speed betweena vending and a magnetic field is given by the general formula;

Frequency 120PN

=

where N = Relative speed between magnetic field and the windingP = Number of poles

For a rotor speed N, the relative speed between the rotating flux and the rotor isNs − N. Consequently, the rotor current frequency f' is given by;

==

−

==

−=

120PNfsf

NNNs120

PNs120

P)NN('f

s

s

ss

s

Q

Q

i.e., Rotor current frequency = Fractional slip x Supply frequency(i) When the rotor is at standstill or stationary (i.e., s = 1), the frequency of

rotor current is the same as that of supply frequency (f' = sf = 1× f = f).

191

(ii) As the rotor picks up speed, the relative speed between the rotating fluxand the rotor decreases. Consequently, the slip s and hence rotor currentfrequency decreases.

Note. The relative speed between the rotating field and stator winding is Ns − 0= Ns. Therefore, the frequency of induced current or voltage in the statorwinding is f = Ns P/120—the supply frequency.

8.8 Effect of Slip on The Rotor CircuitWhen the rotor is stationary, s = 1. Under these conditions, the per phase rotore.m.f. E2 has a frequency equal to that of supply frequency f. At any slip s, therelative speed between stator field and the rotor is decreased. Consequently, therotor e.m.f. and frequency are reduced proportionally to sEs and sf respectively.At the same time, per phase rotor reactance X2, being frequency dependent, isreduced to sX2.

Consider a 6-pole, 3-phase, 50 Hz induction motor. It has synchronous speed Ns

= 120 f/P = 120 × 50/6 = 1000 r.p.m. At standsill, the relative speed betweenstator flux and rotor is 1000 r.p.m. and rotor e.m.f./phase = E2(say). If the full-load speed of the motor is 960 r.p.m., then,

04.010009601000s =−=

(i) The relative speed between stator flux and the rotor is now only 40 r.p.m.Consequently, rotor e.m.f./phase is reduced to:

222 sEorE04.0100040E =×

(ii) The frequency is also reduced in the same ratio to:

sfor04.05010004050 ×=×

(iii) The per phase rotor reactance X2 is likewise reduced to:

222 sXorX04.0100040X =×

Thus at any slip s,Rotor e.m.f./phase = sE2Rotor reactance/phase = sX2Rotor frequency = sf

where E2,X2 and f are the corresponding values at standstill.

192

8.9 Rotor CurrentFig. (8.14) shows the circuit of a 3-phase induction motor at any slip s. The rotoris assumed to be of wound type and star connected. Note that rotor e.m.f./phaseand rotor reactance/phase are s E2 and sX2 respectively. The rotorresistance/phase is R2 and is independent of frequency and, therefore, does notdepend upon slip. Likewise, stator winding values R1 and X1 do not dependupon slip.

Fig.(8.14)

Since the motor represents a balanced 3-phase load, we need consider one phaseonly; the conditions in the other two phases being similar.

At standstill. Fig. (8.15 (i)) shows one phase of the rotor circuit at standstill.

Rotor current/phase,22

22

2

2

22

XR

EZEI

+==

Rotor p.f.,22

22

2

2

22

XR

RZRcos

+==φ

Fig.(8.15)

When running at slip s. Fig. (8.15 (ii)) shows one phase of the rotor circuitwhen the motor is running at slip s.

Rotor current,( )2

222

2

2

22

sXR

sE'Z

sE'I+

==

193

Rotor p.f.,( )2

222

2

2

22

sXR

R'Z

R'cos+

==φ

8.10 Rotor TorqueThe torque T developed by the rotor is directly proportional to:

(i) rotor current(ii) rotor e.m.f.(iii) power factor of the rotor circuit

222 cosIET φ∝∴

or 222 cosIEKT φ=

where I2 = rotor current at standstill E2 = rotor e.m.f. at standstillcos φ2 = rotor p.f. at standstill

Note. The values of rotor e.m.f., rotor current and rotor power factor are takenfor the given conditions.

8.11 Starting Torque (Ts)Let E2 = rotor e.m.f. per phase at standstill

X2 = rotor reactance per phase at standstillR2 = rotor resistance per phase

Rotor impedance/phase, 22

222 XRZ += ...at standstill

Rotor current/phase,22

22

2

2

22

XR

EZEI

+== ...at standstill

Rotor p.f.,22

22

2

2

22

XR

RZRcos

+==φ ...at standstill

∴ Starting torque, 222s cosIEKT φ=

22

22

222

22

22

222

22

22

XRREK

XR

R

XR

EEK

+=

+×

+×=

194

Generally, the stator supply voltage V is constant so that flux per pole φ set upby the stator is also fixed. This in turn means that e.m.f. E2 induced in the rotorwill be constant.

22

2122

22

21s Z

RKXR

RKT =+

=∴

where K1 is another constant.

It is clear that the magnitude of starting torque would depend upon the relativevalues of R2 and X2 i.e., rotor resistance/phase and standstill rotorreactance/phase.

It can be shown that K = 3/2 π Ns.

22

22

222

ss XR

REN2

3T+

⋅π

=∴

Note that here Ns is in r.p.s.

8.12 Condition for Maximum Starting TorqueIt can be proved that starting torque will be maximum when rotorresistance/phase is equal to standstill rotor reactance/phase.

Now 22

22

21s XR

RKT+

= (i)

Differentiating eq. (i) w.r.t. R2 and equating the result to zero, we get,

( )0

XR

)R2(RXR

1KdRdT

222

22

2222

22

12

s =

+−

+=

or 22

22

22 R2XR =+

or 22 XR =

Hence starting torque will be maximum when:Rotor resistance/phase = Standstill rotor reactance/phase

Under the condition of maximum starting torque, φ2 = 45° and rotor powerfactor is 0.707 lagging [See Fig. (8.16 (ii))].

Fig. (8.16 (i)) shows the variation of starting torque with rotor resistance. As therotor resistance is increased from a relatively low value, the starting torqueincreases until it becomes maximum when R2 = X2. If the rotor resistance isincreased beyond this optimum value, the starting torque will decrease.

195

Fig.(8.16)

8.13 Effect of Change of Supply Voltage

22

22

222

s XRREKT

+=

Since E2 ∝ Supply voltage V

22

22

22

2s XR

RVKT+

=∴

where K2 is another constant.2

s VT ∝∴

Therefore, the starting torque is very sensitive to changes in the value of supplyvoltage. For example, a drop of 10% in supply voltage will decrease the startingtorque by about 20%. This could mean the motor failing to start if it cannotproduce a torque greater than the load torque plus friction torque.

8.14 Starting Torque of 3-Phase Induction MotorsThe rotor circuit of an induction motor has low resistance and high inductance.At starting, the rotor frequency is equal to the stator frequency (i.e., 50 Hz) sothat rotor reactance is large compared with rotor resistance. Therefore, rotorcurrent lags the rotor e.m.f. by a large angle, the power factor is low andconsequently the starting torque is small. When resistance is added to the rotorcircuit, the rotor power factor is improved which results in improved startingtorque. This, of course, increases the rotor impedance and, therefore, decreasesthe value of rotor current but the effect of improved power factor predominatesand the starting torque is increased.

(i) Squirrel-cage motors. Since the rotor bars are permanently short-circuited, it is not possible to add any external resistance in the rotor circuitat starting. Consequently, the stalling torque of such motors is low. Squirrel

196

cage motors have starting torque of 1.5 to 2 times the full-load value withstarting current of 5 to 9 times the full-load current.

(ii) Wound rotor motors. The resistance of the rotor circuit of such motorscan be increased through the addition of external resistance. By insertingthe proper value of external resistance (so that R2 = X2), maximum startingtorque can be obtained. As the motor accelerates, the external resistance isgradually cut out until the rotor circuit is short-circuited on itself forrunning conditions.

8.15 Motor Under LoadLet us now discuss the behaviour of 3-phase induction motor on load.

(i) When we apply mechanical load to the shaft of the motor, it will begin toslow down and the rotating flux will cut the rotor conductors at a higherand higher rate. The induced voltage and resulting current in rotorconductors will increase progressively, producing greater and greatertorque.

(ii) The motor and mechanical load will soon reach a state of equilibriumwhen the motor torque is exactly equal to the load torque. When thisstate is reached, the speed will cease to drop any more and the motor willrun at the new speed at a constant rate.

(iii) The drop in speed of the induction motor on increased load is small. It isbecause the rotor impedance is low and a small decrease in speedproduces a large rotor current. The increased rotor current produces ahigher torque to meet the increased load on the motor. This is whyinduction motors are considered to be constant-speed machines.However, because they never actually turn at synchronous speed, theyare sometimes called asynchronous machines.Note that change in load on the induction motor is met through theadjustment of slip. When load on the motor increases, the slip increasesslightly (i.e., motor speed decreases slightly). This results in greaterrelative speed between the rotating flux and rotor conductors.Consequently, rotor current is increased, producing a higher torque tomeet the increased load. Reverse happens should the load on the motordecrease.

(iv) With increasing load, the increased load currents I'2 are in such adirection so as to decrease the stator flux (Lenz’s law), therebydecreasing the counter e.m.f. in the stator windings. The decreasedcounter e.m.f. allows motor stator current (I1) to increase, therebyincreasing the power input to the motor. It may be noted that action ofthe induction motor in adjusting its stator or primary current with

197

changes of current in the rotor or secondary is very much similar to thechanges occurring in transformer with changes in load.

Fig.(8.17)

8.16 Torque Under Running ConditionsLet the rotor at standstill have per phase induced e.m.f. E2, reactance X2 andresistance R2. Then under running conditions at slip s,

Rotor e.m.f./phase, E'2 = sE2

Rotor reactance/phase, X'2 = sX2

Rotor impedance/phase, ( )22

222 sXR'Z +=

Rotor current/phase,( )2

222

2

2

22

sXR

sE'Z'E'I

+==

Rotor p.f.,( )2

222

2m

sXR

R'cos+

=φ

Fig.(8.18)

Running Torque, 222r 'cos'I'ET φ∝

)'E('cos'I 222 φ∝φφ∝ Q

198

( ) ( )

( )

( )

( ))E(

XsRREsK

XsRREsK

XsRREs

XsR

R

XsR

Es

222

22

2221

22

22

22

22

22

22

22

22

22

222

2

φ∝+

=

+

φ=

+

φ∝

+×

+×φ∝

Q

If the stator supply voltage V is constant, then stator flux and hence E2 will beconstant.

( )22

22

22r

XsRRsKT

+=∴

where K2 is another constant.

It may be seen that running torque is:(i) directly proportional to slip i.e., if slip increases (i.e., motor speed

decreases), the torque will increase and vice-versa.(ii) directly proportional to square of supply voltage )VE( 2 ∝Q .

It can be shown that value of K1 = 3/2 π Ns where Ns is in r.p.s.

( ) ( )22

222

s22

22

222

sr

'ZREs

N23

XsRREs

N23T ⋅

π=

+⋅

π=∴

At starting, s = 1 so that starting torque is

22

22

222

ss XR

REN2

3T+

⋅π

=

8.17 Maximum Torque under Running Conditions

22

222

22r XsR

RsKT+

= (i)

In order to find the value of rotor resistance that gives maximum torque underrunning conditions, differentiate exp. (i) w.r.t. s and equate the result to zero i.e.,

( )[ ]( )

0XsR

)Rs(Xs2XsRRKds

dT22

222

2

222

22

22222r =

+

−+=

199

or ( ) 0Xs2XsR 22

22

222 =−+

or 22

222 XsR =

or 22 XsR =

Thus for maximum torque (Tm) under running conditions :

Rotor resistance/phase = Fractional slip × Standstill rotor reactance/phase

Now 22

222

2r XsR

RsT+

∝ … from exp. (i) above

For maximum torque, R2 = s X2. Putting R2 = s X2 in the above expression, themaximum torque Tm is given by;

2m X2

1T ∝

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

It can be shown that:

m-NX2E

N23T

2

22

sm ⋅

π=

It is evident from the above equations that:(i) The value of rotor resistance does not alter the value of the maximum

torque but only the value of the slip at which it occurs.(ii) The maximum torque varies inversely as the standstill reactance.

Therefore, it should be kept as small as possible.(iii) The maximum torque varies directly with the square of the applied

voltage.(iv) To obtain maximum torque at starting (s = 1), the rotor resistance must

be made equal to rotor reactance at standstill.

8.18 Torque-Slip CharacteristicsAs shown in Sec. 8.16, the motor torque under running conditions is given by;

22

222

22

XsRRsKT

+=

If a curve is drawn between the torque and slip for a particular value of rotorresistance R2, the graph thus obtained is called torque-slip characteristic. Fig.(8.19) shows a family of torque-slip characteristics for a slip-range from s = 0 tos = 1 for various values of rotor resistance.

200

Fig.(8.19)

The following points may be noted carefully:(i) At s = 0, T = 0 so that torque-slip curve starts from the origin.(ii) At normal speed, slip is small so that s X2 is negligible as compared to

R2.

2R/sT ∝∴

s∝ ... as R2 is constant

Hence torque slip curve is a straight line from zero slip to a slip thatcorresponds to full-load.

(iii) As slip increases beyond full-load slip, the torque increases and becomesmaximum at s = R2/X2. This maximum torque in an induction motor iscalled pull-out torque or break-down torque. Its value is at least twice thefull-load value when the motor is operated at rated voltage andfrequency.

(iv) to maximum torque, the term22

2Xs increases very rapidly so that 22R may be neglected as compared

to 22

2Xs .22

2 Xs/sT ∝∴

s/1∝ ... as X2 is constant

Thus the torque is now inversely proportional to slip. Hence torque-slipcurve is a rectangular hyperbola.

(v) The maximum torque remains the same and is independent of the valueof rotor resistance. Therefore, the addition of resistance to the rotorcircuit does not change the value of maximum torque but it only changesthe value of slip at which maximum torque occurs.

8.19 Full-Load, Starting and Maximum Torques

201

( )

2m

22

22

2s

22

22

2f

X21T

XRRT

XsRRsT

∝

+∝

+∝

Note that s corresponds to full-load slip.

(i) ( )22

22

22

f

mXRs2XsR

TT +

=∴

Dividing the numerator and denominator on R.H.S. by 22X , we get,

( )( ) sa2

saX/Rs2

sX/RTT 22

22

2222

f

m +=+

=

where phasereactance/rotorStandstill/phaseresistanceRotor

XRa

2

2 ==

(ii)22

22

22

s

mXR2XR

TT +

=

Dividing the numerator and denominator on R.H.S. by 22X , we get,

( )( ) a2

1aX/R2

1X/RTT 2

22

222

f

m +=+

=

where phasereactance/rotorStandstill/phaseresistanceRotor

XRa

2

2 ==

8.20 Induction Motor and Transformer ComparedAn induction motor may be considered to be a transformer with a rotating short-circuited secondary. The stator winding corresponds to transformer primary androtor winding to transformer secondary. However, the following differencesbetween the two are worth noting:(i) Unlike a transformer, the magnetic circuit of a 3-phase induction motor has

an air gap. Therefore, the magnetizing current in a 3-phase induction motoris much larger than that of the transformer. For example, in an inductionmotor, it may be as high as 30-50 % of rated current whereas it is only 1-5% of rated current in a transformer.

(ii) In an induction motor, there is an air gap and the stator and rotor windingsare distributed along the periphery of the air gap rather than concentrated

202

on a core as in a transformer. Therefore, the leakage reactances of statorand rotor windings are quite large compared to that of a transformer.

(iii) In an induction motor, the inputs to the stator and rotor are electrical but theoutput from the rotor is mechanical. However, in a transformer, input aswell as output is electrical.

(iv) The main difference between the induction motor and transformer lies inthe fact that the rotor voltage and its frequency are both proportional to slips. If f is the stator frequency, E2 is the per phase rotor e.m.f. at standstilland X2 is the standstill rotor reactance/phase, then at any slip s, these valuesare:

Rotor e.m.f./phase, E'2 = s E2

Rotor reactance/phase, X'2 = sX2

Rotor frequency, f' = sf

8.21 Speed Regulation of Induction MotorsLike any other electrical motor, the speed regulation of an induction motor isgiven by:

% age speed regulation 100NNN

.L.F

.L.F0 ×−

=

where N0 = no-load speed of the motorNF.L. = full-load speed of the motor

If the no-load speed of the motor is 800 r.p.m. and its fall-load speed in 780r.p.m., then change in speed is 800 − 780 = 20 r.p.m. and percentage speedregulation = 20 × 100/780 = 2.56%.

At no load, only a small torque is required to overcome the small mechanicallosses and hence motor slip is small i.e., about 1%. When the motor is fullyloaded, the slip increases slightly i.e., motor speed decreases slightly. It isbecause rotor impedance is low and a small decrease in speed produces a largerotor current. The increased rotor current produces a high torque to meet the fullload on the motor. For this reason, the change in speed of the motor from no-load to full-load is small i.e., the speed regulation of an induction motor is low.The speed regulation of an induction motor is 3% to 5%. Although the motorspeed does decrease slightly with increased load, the speed regulation is lowenough that the induction motor is classed as a constant-speed motor.

8.22 Speed Control of 3-Phase Induction Motors

203

Pf120)s1(N)s1(N s −=−= (i)

An inspection of eq. (i) reveals that the speed N of an induction motor can bevaried by changing (i) supply frequency f (ii) number of poles P on the statorand (iii) slip s. The change of frequency is generally not possible because thecommercial supplies have constant frequency. Therefore, the practical methodsof speed control are either to change the number of stator poles or the motor slip.

1. Squirrel cage motorsThe speed of a squirrel cage motor is changed by changing the number of statorpoles. Only two or four speeds are possible by this method. Two-speed motorhas one stator winding that may be switched through suitable control equipmentto provide two speeds, one of which is half of the other. For instance, thewinding may be connected for either 4 or 8 poles, giving synchronous speeds of1500 and 750 r.p.m. Four-speed motors are equipped with two separate statorwindings each of which provides two speeds. The disadvantages of this methodare:

(i) It is not possible to obtain gradual continuous speed control.(ii) Because of the complications in the design and switching of the

interconnections of the stator winding, this method can provide amaximum of four different synchronous speeds for any one motor.

2. Wound rotor motorsThe speed of wound rotor motors is changed by changing the motor slip. Thiscan be achieved by;

(i) varying the stator line voltage(ii) varying the resistance of the rotor circuit(iii) inserting and varying a foreign voltage in the rotor circuit

8.23 Power Factor of Induction MotorLike any other a.c. machine, the power factor of an induction motor is given by;

Power factor, cos (I)currentTotal)cos(IcurrentofcomponentActive φ

=φ

The presence of air-gap between the stator and rotor of an induction motorgreatly increases the reluctance of the magnetic circuit. Consequently, aninduction motor draws a large magnetizing current (Im) to produce the requiredflux in the air-gap.

(i) At no load, an induction motor draws a large magnetizing current and asmall active component to meet the no-load losses. Therefore, theinduction motor takes a high no-load current lagging the applied voltage

204

by a large angle. Hence the power factor of an induction motor on noload is low i.e., about 0.1 lagging.

(ii) When an induction motor is loaded, the active component of currentincreases while the magnetizing component remains about the same.Consequently, the power factor of the motor is increased. However,because of the large value of magnetizing current, which is presentregardless of load, the power factor of an induction motor even at full-load seldom exceeds 0.9 lagging.

8.24 Power Stages in an Induction MotorThe input electric power fed to the stator of the motor is converted intomechanical power at the shaft of the motor. The various losses during the energyconversion are:

1. Fixed losses(i) Stator iron loss(ii) Friction and windage loss

The rotor iron loss is negligible because the frequency of rotor currents undernormal running condition is small.

2. Variable losses(i) Stator copper loss(ii) Rotor copper loss

Fig. (8.20) shows how electric power fed to the stator of an induction motorsuffers losses and finally converted into mechanical power.

The following points may be noted from the above diagram:(i) Stator input, Pi = Stator output + Stator losses

= Stator output + Stator Iron loss + Stator Cu loss(ii) Rotor input, Pr = Stator output

It is because stator output is entirely transferred to the rotor through air-gap by electromagnetic induction.

(iii) Mechanical power available, Pm = Pr − Rotor Cu lossThis mechanical power available is the gross rotor output and willproduce a gross torque Tg.

(iv) Mechanical power at shaft, Pout = Pm − Friction and windage lossMechanical power available at the shaft produces a shaft torque Tsh.

Clearly, Pm − Pout = Friction and windage loss

205

Fig.(8.20)

8.25 Induction Motor TorqueThe mechanical power P available from any electric motor can be expressed as:

watts60TN2P π

=

where N = speed of the motor in r.p.m.T = torque developed in N-m

m-NNP55.9N

P260T =π

=∴

If the gross output of the rotor of an induction motor is Pm and its speed is Nr.p.m., then gross torque T developed is given by:

m-NNP55.9T m

g =

Similarly, m-NNP55.9T out

sh =

Note. Since windage and friction loss is small, Tg = Tsh,. This assumption hardlyleads to any significant error.

8.26 Rotor OutputIf Tg newton-metre is the gross torque developed and N r.p.m. is the speed of therotor, then,

Gross rotor output = 60TN2 gπ

watts

If there were no copper losses in the rotor, the output would equal rotor inputand the rotor would run at synchronous speed Ns.

206

∴ Rotor input = 60TN2 gsπ

watts

∴ Rotor Cu loss = Rotor input − Rotor output

)NN(60T2

sg −

π=

(i) sNNN

inputRotorlossCuRotor

s

s =−

=

∴ Rotor Cu loss = s × Rotor input

(ii) Gross rotor output, Pm = Rotor input − Rotor Cu loss = Rotor input − s × Rotor input

∴ Pm = Rotor input (1 − s)

(iii)sN

Ns1inputRotoroutputrotorGross

=−=

(iv) s1s

outputrotorGrosslossCuRotor

−=

It is clear that if the input power to rotor is Pr then s Pr is lost as rotor Cu lossand the remaining (1 − s)Pr is converted into mechanical power. Consequently,induction motor operating at high slip has poor efficiency.

Note.

s1inputRotoroutputrotorGross

−=

If the stator losses as well as friction and windage losses arc neglected, then, Gross rotor output = Useful output

Rotor input = Stator input

Efficiencys1outputStatoroutputUseful

=−=∴

Hence the approximate efficiency of an induction motor is 1 − s. Thus if the slipof an induction motor is 0.125, then its approximate efficiency is = 1 − 0.125 =0.875 or 87.5%.

8.27 Induction Motor Torque EquationThe gross torque Tg developed by an induction motor is given by;

207

r.p.s.isN...8.26)Sec.(SeeN2inputRotor60

r.p.s.isN...N2inputRotorT

ss

ss

g

π×=

π=

Now Rotor input ( )s

R'I3s

lossCuRotor 22

2== (i)

As shown in Sec. 8.16, under running conditions,

22

22

12

222

22

)Xs(R

EKs

)Xs(R

Es'I+

=+

=

where K = Transformation ratio = ns/phaseStator turs/phaseRotor turn

∴ Rotor input = 22

22

222

22

22

222

2

)Xs(RREs3

s1

)Xs(RREs3

+=×

+×

(Putting me value of I'2 in eq.(i))

Also Rotor input = 22

22

221

2

22

22

221

22

)Xs(RREKs3

s1

)Xs(RREKs3

+=×

+×

(Putting me value of I'2 in eq.(i))

22

22

222

ssg )Xs(R

REsN2

3N2inputRotorT

+×

π=

π=∴ ...in terms of E2

22

22

221

2

s )Xs(RREKs

N23

+×

π= ...in terms of E1

Note that in the above expressions of Tg, the values E1, E2, R2 and X2 representthe phase values.

8.28 Performance Curves of Squirrel-Cage MotorThe performance curves of a 3-phase induction motor indicate the variations ofspeed, power factor, efficiency, stator current and torque for different values ofload. However, before giving the performance curves in one graph, it isdesirable to discuss the variation of torque, and stator current with slip.

(i) Variation of torque and stator current with slipFig. (8.21) shows the variation of torque and stator current with slip for astandard squirrel-cage motor. Generally, the rotor resistance is low so that full-

208

load current occurs at low slip. Then even at full-load f' (= sf) and. therefore, X'2(= 2π f' L2) are low. Between zero and full-load, rotor power factor (= cos φ'2)and rotor impedance (= Z'2) remain practically constant. Therefore, rotor currentI'2(E'2/Z'2) and, therefore, torque (Tr) increase directly with the slip. Now statorcurrent I1 increases in proportion to I'2. This is shown in Fig. (8.21) where Tr andI1 are indicated as straight lines from no-load to full-load. As load and slip areincreased beyond full-load, the increase in rotor reactance becomes appreciable.The increasing value of rotor impedance not only decreases the rotor powerfactor cos φ'2 (= R2/Z'2) but also lowers the rate of increase of rotor current. As aresult, the torque Tr and stator current I1 do not increase directly with slip asindicated in Fig. (8.21). With the decreasing power factor and the lowered rateof increase in rotor current, the stator current I1 and torque Tr increase at a lowerrate. Finally, torque Tr reaches the maximum value at about 25% slip in thestandard squirrel cage motor. This maximum value of torque is called thepullout torque or breakdown torque. If the load is increased beyond thebreakdown point, the decrease in rotor power factor is greater than the increasein rotor current, resulting in a decreasing torque. The result is that motor slowsdown quickly and comes to a stop.

Fig.(8.21)

In Fig. (8.21), the value of torque at starting (i.e., s = 100%) is 1.5 times the full-load torque. The starting current is about five times the full-load current. Themotor is essentially a constant-speed machine having speed characteristics aboutthe same as a d.c. shunt motor.

(ii) Performance curvesFig. (8.22) shows the performance curves of 3-phase squirrel cage inductionmotor.

209

Fig.(8.22)

The following points may be noted:(a) At no-load, the rotor lags behind the stator flux by only a small amount,

since the only torque required is that needed to overcome the no-loadlosses. As mechanical load is added, the rotor speed decreases. A decreasein rotor speed allows the constant-speed rotating field to sweep across therotor conductors at a faster rate, thereby inducing large rotor currents. Thisresults in a larger torque output at a slightly reduced speed. This explainsfor speed-load curve in Fig. (8.22).

(b) At no-load, the current drawn by an induction motor is largely amagnetizing current; the no-load current lagging the applied voltage by alarge angle. Thus the power factor of a lightly loaded induction motor isvery low. Because of the air gap, the reluctance of the magnetic circuit ishigh, resulting in a large value of no-load current as compared with atransformer. As load is added, the active or power component of currentincreases, resulting in a higher power factor. However, because of the largevalue of magnetizing current, which is present regardless of load, the powerfactor of an induction motor even at full-load seldom exceeds 90%. Fig.(8.22) shows the variation of power factor with load of a typical squirrel-cage induction motor.

(c) Efficiency LossesOutputOutput

+=

The losses occurring in a 3-phase induction motor are Cu losses in statorand rotor windings, iron losses in stator and rotor core and friction andwindage losses. The iron losses and friction and windage losses are almostindependent of load. Had I2R been constant, the efficiency of the motorwould have increased with load. But I2R loss depends upon load.

210

Therefore, the efficiency of the motor increases with load but the curve isdropping at high loads.

(d) At no-load, the only torque required is that needed to overcome no-loadlosses. Therefore, stator draws a small current from the supply. Asmechanical load is added, the rotor speed decreases. A decrease in rotorspeed allows the constant-speed rotating field to sweep across the rotorconductors at a faster rate, thereby inducing larger rotor currents. Withincreasing loads, the increased rotor currents are in such a direction so as todecrease the staler flux, thereby temporarily decreasing the counter e.m.f.in the stator winding. The decreased counter e.m.f. allows more statorcurrent to flow.

(e) Output = Torque × SpeedSince the speed of the motor does not change appreciably with load, thetorque increases with increase in load.

8.29 Equivalent Circuit of 3-Phase Induction Motor atAny Slip

In a 3-phase induction motor, the stator winding is connected to 3-phase supplyand the rotor winding is short-circuited. The energy is transferred magneticallyfrom the stator winding to the short-circuited, rotor winding. Therefore, aninduction motor may be considered to be a transformer with a rotating secondary(short-circuited). The stator winding corresponds to transformer primary and therotor finding corresponds to transformer secondary. In view of the similarity ofthe flux and voltage conditions to those in a transformer, one can expect that theequivalent circuit of an induction motor will be similar to that of a transformer.Fig. (8.23) shows the equivalent circuit (though not the only one) per phase foran induction motor. Let us discuss the stator and rotor circuits separately.

Fig.(8.23)

Stator circuit. In the stator, the events are very similar to those in thetransformer primal y. The applied voltage per phase to the stator is V1 and R1and X1 are the stator resistance and leakage reactance per phase respectively.The applied voltage V1 produces a magnetic flux which links the stator winding(i.e., primary) as well as the rotor winding (i.e., secondary). As a result, self-

211

Fig.(8.24)

induced e.m.f. E1 is induced in the stator winding and mutually induced e.m.f.E'2(= s E2 = s K E1 where K is transformation ratio) is induced in the rotorwinding. The flow of stator current I1 causes voltage drops in R1 and X1.

)XjR(IEV 11111 ++−=∴ ...phasor sum

When the motor is at no-load, the stator winding draws a current I0. It has twocomponents viz., (i) which supplies the no-load motor losses and (ii)magnetizing component Im which sets up magnetic flux in the core and the air-gap. The parallel combination of Rc and Xm, therefore, represents the no-loadmotor losses and the production of magnetic flux respectively.

mw0 III +=

Rotor circuit. Here R2 and X2 represent the rotor resistance and standstill rotorreactance per phase respectively. At any slip s, the rotor reactance will be s X2The induced voltage/phase in the rotor is E'2 = s E2 = s K E1. Since the rotorwinding is short-circuited, the whole of e.m.f. E'2 is used up in circulating therotor current I'2.

)XsjR('I'E 2222 +=∴

The rotor current I'2 is reflected as I"2 (= K I'2) in the stator. The phasor sum ofI"2 and I0 gives the stator current I1.

It is important to note that input to the primary and output from the secondary ofa transformer are electrical. However, in an induction motor, the inputs to thestator and rotor are electrical but the output from the rotor is mechanical. Tofacilitate calculations, it is desirable andnecessary to replace the mechanical load by anequivalent electrical load. We then have thetransformer equivalent circuit of the inductionmotor.

It may be noted that even though the frequenciesof stator and rotor currents are different, yet themagnetic fields due to them rotate atsynchronous speed Ns. The stator currentsproduce a magnetic flux which rotates at a speedNs. At slip s, the speed of rotation of the rotorfield relative to the rotor surface in the directionof rotation of the rotor is

sNsPfs120

P'f120

===

But the rotor is revolving at a speed of N relativeto the stator core. Therefore, the speed of rotor

212

field relative to stator core

sss NN)NN(NsN =+−=+=

Thus no matter what the value of slip s, the stator and rotor magnetic fields aresynchronous with each other when seen by an observer stationed in space.Consequently, the 3-phase induction motor can be regarded as being equivalentto a transformer having an air-gap separating the iron portions of the magneticcircuit carrying the primary and secondary windings.

Fig. (8.24) shows the phasor diagram of induction motor.

8.30 Equivalent Circuit of the RotorWe shall now see how mechanical load of the motor is replaced by theequivalent electrical load. Fig. (8.25 (i)) shows the equivalent circuit per phaseof the rotor at slip s. The rotor phase current is given by;

22

22

22

)Xs(R

Es'I+

=

Mathematically, this value is unaltered by writing it as:

22

22

22

)X()s/R(

E'I+

=

As shown in Fig. (8.25 (ii)), we now have a rotor circuit that has a fixedreactance X2 connected in series with a variable resistance R2/s and suppliedwith constant voltage E2. Note that Fig. (8.25 (ii)) transfers the variable to theresistance without altering power or power factor conditions.

Fig.(8.25)

The quantity R2/s is greater than R2 since s is a fraction. Therefore, R2/s can bedivided into a fixed part R2 and a variable part (R2/s − R2) i.e.,

−+= 1s1RRs

R22

2

213

(i) The first part R2 is the rotor resistance/phase, and represents the rotor Culoss.

(ii) The second part

−1s1R2 is a variable-resistance load. The power

delivered to this load represents the total mechanical power developed inthe rotor. Thus mechanical load on the induction motor can be replaced by

a variable-resistance load of value

−1s1R2 . This is

−=∴ 1s1RR 2L

Fig. (8.25 (iii)) shows the equivalent rotor circuit along with load resistance RL.

8.31 Transformer Equivalent Circuit of Induction MotorFig. (8.26) shows the equivalent circuit per phase of a 3-phase induction motor.Note that mechanical load on the motor has been replaced by an equivalentelectrical resistance RL given by;

−= 1s1RR 2L (i)

Fig.(8.26)

Note that circuit shown in Fig. (8.26) is similar to the equivalent circuit of atransformer with secondary load equal to R2 given by eq. (i). The rotor e.m.f. inthe equivalent circuit now depends only on the transformation ratio K (= E2/E1).

Therefore; induction motor can be represented as an equivalent transformerconnected to a variable-resistance load RL given by eq. (i). The power deliveredto RL represents the total mechanical power developed in the rotor. Since theequivalent circuit of Fig. (8.26) is that of a transformer, the secondary (i.e.,rotor) values can be transferred to primary (i.e., stator) through the appropriateuse of transformation ratio K. Recall that when shifting resistance/reactancefrom secondary to primary, it should be divided by K2 whereas current should bemultiplied by K. The equivalent circuit of an induction motor referred toprimary is shown in Fig. (8.27).

214

Fig.(8.27)

Note that the element (i.e., R'L) enclosed in the dotted box is the equivalentelectrical resistance related to the mechanical load on the motor. The followingpoints may be noted from the equivalent circuit of the induction motor:(i) At no-load, the slip is practically zero and the load R'L is infinite. This

condition resembles that in a transformer whose secondary winding isopen-circuited.

(ii) At standstill, the slip is unity and the load R'L is zero. This conditionresembles that in a transformer whose secondary winding is short-circuited.

(iii) When the motor is running under load, the value of R'L will depend uponthe value of the slip s. This condition resembles that in a transformer whosesecondary is supplying variable and purely resistive load.

(iv) The equivalent electrical resistance R'L related to mechanical load is slip orspeed dependent. If the slip s increases, the load R'L decreases and the rotorcurrent increases and motor will develop more mechanical power. This isexpected because the slip of the motor increases with the increase of loadon the motor shaft.

8.32 Power RelationsThe transformer equivalent circuit of an induction motor is quite helpful inanalyzing the various power relations in the motor. Fig. (8.28) shows theequivalent circuit per phase of an induction motor where all values have beenreferred to primary (i.e., stator).

Fig.(8.28)

215

(i) Total electrical load s'R'R1s

1'R 222 =+

−=

Power input to stator = 111 cosIV3 φ

There will be stator core loss and stator Cu loss. The remaining power willbe the power transferred across the air-gap i.e., input to the rotor.

(ii) Rotor input = ( )s

'R"I3 22

2

Rotor Cu loss = ( ) 22

2 'R"I3

Total mechanical power developed by the rotor is

Pm = Rotor input − Rotor Cu loss

= ( ) ( ) ( )

−= 1s1'R"I3'R"I3s

'R"I32

222

22

22

2

This is quite apparent from the equivalent circuit shown in Fig. (8.28).

(iii) If Tg is the gross torque developed by the rotor, then,

60TN2

P gm

π=

or ( ) 60TN2

1s1'R"I3 g

22

2π

=

−

or ( ) 60TN2

ss1'R"I3 g

22

2π

=

−

or ( ) [ ])s1(NN60T)s1(N2

ss1'R"I3 s

gs2

22 −=

−π=

−

Q

( ) m-N60N2s'R"I3T

s

22

2g π

=∴

or ( ) m-NNs'R"I355.9T

s

22

2g =

Note that shaft torque Tsh will be less than Tg by the torque required to meetwindage and frictional losses.

216

8.33 Approximate Equivalent Circuit of Induction MotorAs in case of a transformer, the approximate equivalent circuit of an inductionmotor is obtained by shifting the shunt branch (Rc − Xm) to the input terminalsas shown in Fig. (8.29). This step has been taken on the assumption that voltagedrop in R1 and X1 is small and the terminal voltage V1 does not appreciablydiffer from the induced voltage E1. Fig. (8.29) shows the approximate equivalentcircuit per phase of an induction motor where all values have been referred toprimary (i.e., stator).

Fig.(8.29)

The above approximate circuit of induction motor is not so readily justified aswith the transformer. This is due to the following reasons:(i) Unlike that of a power transformer, the magnetic circuit of the induction

motor has an air-gap. Therefore, the exciting current of induction motor (30to 40% of full-load current) is much higher than that of the powertransformer. Consequently, the exact equivalent circuit must be used foraccurate results.

(ii) The relative values of X1 and X2 in an induction motor are larger than thecorresponding ones to be found in the transformer. This fact does notjustify the use of approximate equivalent circuit

(iii) In a transformer, the windings are concentrated whereas in an inductionmotor, the windings are distributed. This affects the transformation ratio.

In spite of the above drawbacks of approximate equivalent circuit, it yieldsresults that are satisfactory for large motors. However, approximate equivalentcircuit is not justified for small motors.

8.34 Starting of 3-Phase Induction MotorsThe induction motor is fundamentally a transformer in which the stator is theprimary and the rotor is short-circuited secondary. At starting, the voltageinduced in the induction motor rotor is maximum (Q s = 1). Since the rotorimpedance is low, the rotor current is excessively large. This large rotor currentis reflected in the stator because of transformer action. This results in highstarting current (4 to 10 times the full-load current) in the stator at low power

217

factor and consequently the value of starting torque is low. Because of the shortduration, this value of large current does not harm the motor if the motoraccelerates normally. However, this large starting current will produce largeline-voltage drop. This will adversely affect the operation of other electricalequipment connected to the same lines. Therefore, it is desirable and necessaryto reduce the magnitude of stator current at starting and several methods areavailable for this purpose.

8.35 Methods of Starting 3-Phase Induction MotorsThe method to be employed in starting a given induction motor depends uponthe size of the motor and the type of the motor. The common methods used tostart induction motors are:

(i) Direct-on-line starting (ii) Stator resistance starting(iii) Autotransformer starting (iv) Star-delta starting(v) Rotor resistance starting

Methods (i) to (iv) are applicable to both squirrel-cage and slip ring motors.However, method (v) is applicable only to slip ring motors. In practice, any oneof the first four methods is used for starting squirrel cage motors, dependingupon ,the size of the motor. But slip ring motors are invariably started by rotorresistance starting.

8.36 Methods of Starting Squirrel-Cage MotorsExcept direct-on-line starting, all other methods of starting squirrel-cage motorsemploy reduced voltage across motor terminals at starting.

(i) Direct-on-line startingThis method of starting in just what the name implies—the motor is started byconnecting it directly to 3-phase supply. The impedance of the motor atstandstill is relatively low and when it is directly connected to the supplysystem, the starting current will be high (4 to 10 times the full-load current) andat a low power factor. Consequently, this method of starting is suitable forrelatively small (up to 7.5 kW) machines.

Relation between starling and F.L. torques. We know that:

Rotor input = 2π Ns T = kT

But Rotor Cu loss = s × Rotor input

( ) kTsR'I3 22

2 ×=∴

or ( ) s'IT 22∝

218

or )I'I(sIT 1221 ∝∝ Q

If Ist is the starting current, then starting torque (Tst) is)1sstartingat(IT 2

st =∝ Q

If If is the full-load current and sf is the full-load slip, then,

f

2

f

st

f

st

f2ff

sII

TT

sIT

×

=∴

∝

When the motor is started direct-on-line, the starting current is the short-circuit(blocked-rotor) current Isc.

f

2

f

sc

f

st sII

TT

×

=∴

Let us illustrate the above relation with a numerical example. Suppose Isc = 5 Ifand full-load slip sf =0.04. Then,

fst

22

f

ff

2

f

sc

f

st

TT

104.0)5(04.0II5sI

ITT

=∴

=×=×

=×

=

Note that starting current is as large as five times the full-load current butstarting torque is just equal to the full-load torque. Therefore, starting current isvery high and the starting torque is comparatively low. If this large startingcurrent flows for a long time, it may overheat the motor and damage theinsulation.

(ii) Stator resistance startingIn this method, external resistances are connected in series with each phase ofstator winding during starting. This causes voltage drop across the resistances sothat voltage available across motor terminals is reduced and hence the startingcurrent. The starting resistances are gradually cut out in steps (two or moresteps) from the stator circuit as the motor picks up speed. When the motorattains rated speed, the resistances are completely cut out and full line voltage isapplied to the rotor.

This method suffers from two drawbacks. First, the reduced voltage applied tothe motor during the starting period lowers the starting torque and henceincreases the accelerating time. Secondly, a lot of power is wasted in the startingresistances.

219

Fig.(8.30)

Relation between starting and F.L. torques. Let V be the rated voltage/phase.If the voltage is reduced by a fraction x by the insertion of resistors in the line,then voltage applied to the motor per phase will be xV.

scst IxI =

Now f

2

f

st

f

st sII

TT

×

=

or f

2

f

sc2

f

st sIIxT

T×

=

Thus while the starting current reduces by a fraction x of the rated-voltagestarting current (Isc), the starting torque is reduced by a fraction x2 of thatobtained by direct switching. The reduced voltage applied to the motor duringthe starting period lowers the starting current but at the same time increases theaccelerating time because of the reduced value of the starting torque. Therefore,this method is used for starting small motors only.

(iii) Autotransformer startingThis method also aims at connecting the induction motor to a reduced supply atstarting and then connecting it to the full voltage as the motor picks up sufficientspeed. Fig. (8.31) shows the circuit arrangement for autotransformer starting.The tapping on the autotransformer is so set that when it is in the circuit, 65% to80% of line voltage is applied to the motor.

At the instant of starting, the change-over switch is thrown to “start” position.This puts the autotransformer in the circuit and thus reduced voltage is appliedto the circuit. Consequently, starting current is limited to safe value. When themotor attains about 80% of normal speed, the changeover switch is thrown to

220

“run” position. This takes out the autotransformer from the circuit and puts themotor to full line voltage. Autotransformer starting has several advantages vizlow power loss, low starting current and less radiated heat. For large machines(over 25 H.P.), this method of starting is often used. This method can be usedfor both star and delta connected motors.

Fig.(8.31)

Relation between starting And F.L. torques. Consider a star-connectedsquirrel-cage induction motor. If V is the line voltage, then voltage across motorphase on direct switching is 3V and starting current is Ist = Isc. In case ofautotransformer, if a tapping of transformation ratio K (a fraction) is used, thenphase voltage across motor is 3KV and Ist = K Isc,

Now f

2

f

sc2f

2

f

scf

2

f

st

f

st sIIKsI

IKsII

TT

×

=×

=×

=

f

2

f

sc2

f

st sIIKT

T×

=∴

Fig.(8.32)

221

The current taken from the supply or by autotransformer is I1 = KI2 = K2Isc. Notethat motor current is K times, the supply line current is K2 times and the startingtorque is K2 times the value it would have been on direct-on-line starting.

(iv) Star-delta startingThe stator winding of the motor is designed for delta operation and is connectedin star during the starting period. When the machine is up to speed, theconnections are changed to delta. The circuit arrangement for star-delta startingis shown in Fig. (8.33).

The six leads of the stator windings are connected to the changeover switch asshown. At the instant of starting, the changeover switch is thrown to “Start”position which connects the stator windings in star. Therefore, each stator phasegets 3V volts where V is the line voltage. This reduces the starting current.When the motor picks up speed, the changeover switch is thrown to “Run”position which connects the stator windings in delta. Now each stator phase getsfull line voltage V. The disadvantages of this method are:(a) With star-connection during starting, stator phase voltage is 31 times the

line voltage. Consequently, starting torque is ( )231 or 1/3 times the valueit would have with ∆-connection. This is rather a large reduction in startingtorque.

(b) The reduction in voltage is fixed.

This method of starting is used for medium-size machines (upto about 25 H.P.).

Relation between starting and F.L. torques. In direct delta starting,Starting current/phase, Isc = V/Zsc where V = line voltage

Starting line current = scI3In star starting, we have,

Starting current/phase, scsc

st I3

1Z

3VI ==

Now f

2

f

scf

2

f

st

f

st sI3

IsII

TT

×

×=×

=

or f

2

f

sc

f

st sII

31

TT

×

=

where Isc = starting phase current (delta) If = F.L. phase current (delta)

222

Fig.(8.33)

Note that in star-delta starting, the starting line current is reduced to one-third ascompared to starting with the winding delta connected. Further, starting torqueis reduced to one-third of that obtainable by direct delta starting. This method ischeap but limited to applications where high starting torque is not necessary e.g.,machine tools, pumps etc.

8.37 Starting of Slip-Ring MotorsSlip-ring motors are invariably started by rotor resistance starting. In thismethod, a variable star-connected rheostat is connected in the rotor circuitthrough slip rings and full voltage is applied to the stator winding as shown inFig. (8.34).

Fig.(8.34)

(i) At starting, the handle of rheostat is set in the OFF position so thatmaximum resistance is placed in each phase of the rotor circuit. Thisreduces the starting current and at the same time starting torque isincreased.

(ii) As the motor picks up speed, the handle of rheostat is gradually moved inclockwise direction and cuts out the external resistance in each phase of therotor circuit. When the motor attains normal speed, the change-over switchis in the ON position and the whole external resistance is cut out from therotor circuit.

223

8.38 Slip-Ring Motors Versus Squirrel Cage MotorsThe slip-ring induction motors have the following advantages over the squirrelcage motors:

(i) High starting torque with low starting current.(ii) Smooth acceleration under heavy loads.(iii) No abnormal heating during starting.(iv) Good running characteristics after external rotor resistances are cut out.(v) Adjustable speed.

The disadvantages of slip-ring motors are:(i) The initial and maintenance costs are greater than those of squirrel cage

motors.(ii) The speed regulation is poor when run with resistance in the rotor circuit

8.39 Induction Motor RatingThe nameplate of a 3-phase induction motor provides the following information:

(i) Horsepower (ii) Line voltage (iii) Line current(iv) Speed (v) Frequency (vi) Temperature rise

The horsepower rating is the mechanical output of the motor when it is operatedat rated line voltage, rated frequency and rated speed. Under these conditions,the line current is that specified on the nameplate and the temperature rise doesnot exceed that specified.

The speed given on the nameplate is the actual speed of the motor at rated full-load; it is not the synchronous speed. Thus, the nameplate speed of the inductionmotor might be 1710 r.p.m. It is the rated full-load speed.

8.40 Double Squirrel-Cage MotorsOne of the advantages of the slip-ring motor is that resistance may be inserted inthe rotor circuit to obtain high starting torque (at low starting current) and thencut out to obtain optimum running conditions. However, such a procedurecannot be adopted for a squirrel cage motor because its cage is permanentlyshort-circuited. In order to provide high starting torque at low starting current,double-cage construction is used.

ConstructionAs the name suggests, the rotor of this motor has two squirrel-cage windingslocated one above the other as shown in Fig. (8.35 (i)).(i) 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

224

outer winding has relatively open slots and a poorer flux path around itsbars [See Fig. (8.35 (ii))], it has a low inductance. Thus the resistance ofthe outer squirrel-cage winding is high and its inductance is low.

(ii) The inner winding consists of bars of greater cross-section short-circuitedby end rings. Therefore, the resistance of this winding is low. Since thebars of the inner winding are thoroughly buried in iron, it has a highinductance [See Fig. (8.35 (ii))]. Thus the resistance of the inner squirrel-cage winding is low and its inductance is high.

Fig.(8.35)

WorkingWhen a rotating magnetic field sweeps across the two windings, equal e.m.f.sare induced in each.(i) 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 theupper winding. Because of the high reactance of the lower winding, nearlyall the rotor current flows in the high-resistance outer cage winding. Thisprovides the good starting characteristics of a high-resistance cage winding.Thus the outer winding gives high starting torque at low starting current.

(ii) As the motor accelerates, the rotor frequency decreases, thereby loweringthe reactance of the inner winding, allowing it to carry a larger proportionof the total rotor current At the normal operating speed of the motor, therotor frequency is so low (2 to 3 Hz) that nearly all the rotor current flowsin the low-resistance inner cage winding. This results in good operatingefficiency and speed regulation.

Fig. (8.36) shows the operating characteristics of double squirrel-cage motor.The starting torque of this motor ranges from 200 to 250 percent of full-loadtorque with a starting current of 4 to 6 times the full-load value. It is classed as ahigh-torque, low starting current motor.

225

Fig.(8.37)

Fig.(8.36)

8.41 Equivalent Circuit of Double Squirrel-Cage MotorFig. (8.37) shows a section of the double squirrel cage motor.Here Ro and Ri are the per phase resistances of the outer cagewinding and inner cage winding whereas Xo and Xi are thecorresponding per phase standstill reactances. For the outercage, the resistance is made intentionally high, giving a highstarting torque. For the inner cage winding, the resistance is lowand the leakage reactance is high, giving a low starting torquebut high efficiency on load. Note that in a double squirrel cage motor, the outerwinding produces the high starting and accelerating torque while the innerwinding provides the running torque at good efficiency.

Fig. (8.38 (i)) shows the equivalent circuit for one phase of double cage motorreferred to stator. The two cage impedances are effectively in parallel. Theresistances and reactances of the outer and inner rotors are referred to the stator.The exciting circuit is accounted for as in a single cage motor. If themagnetizing current (I0) is neglected, then the circuit is simplified to that shownin Fig. (8.38 (ii)).

Fig.(8.38)

226

From the equivalent circuit, the performance of the motor can be predicted.

Total impedance as referred to stator is

oi

oi11

oi111o 'Z'Z

'Z'ZXjR'Z1'Z11XjRZ

+++=

+++=

1

Tests for finding parameters of 3Φ induction motor

To accurately model the three phase induction motor , the

equivalent circuit parameters (R1 , X1 , R2 , X2 , Rc , Xm)

should be known. A single-phase equivalent circuit with lumped

parameters is the most traditional model of an AC induction

motor .The steady-state operating characteristics of a three-phase

induction motor are often investigated using a per-phase

equivalent circuit . In this circuit R1, and X1 represent stator

resistance and leakage reactance, respectively; R2 and X2 denote

the rotor resistance and leakage reactance referred to the stator,

respectively; Rc resistance stands for core losses; Xm represents

magnetizing reactance. The equivalent circuit is used to facilitate

the computation of various operating quantities, such as stator

current, input power, losses, induced torque, and efficiency.

When power aspects of the operation need to be emphasized, the

shunt resistance is usually neglected; the core losses can be

included in efficiency calculations along with the friction,

windage, and stray losses.

The parameters of the equivalent circuit can be obtained from the

dc, no-load, and blocked-rotor tests .The DC Test is performed to

compute the stator winding resistance . A dc voltage is applied to

the stator windings of an induction motor. The resulting current

flowing through the stator windings is a dc current; thus, no

voltage is induced in the rotor circuit, and the motor reactance is

zero. The stator resistance is the only circuit parameter limiting

current flow. In No-Load Test , a rated, balanced ac voltage at a

rated frequency is applied to the stator while it is running at no

load, and input power, voltage, and phase currents are measured at

the no-load condition. In Blocked-Rotor Test, the rotor of the

induction motor is blocked, and a reduced voltage is applied to the

stator terminals so that the rated current flows through the stator

windings. The input power, voltage, and current are measured.

6

Notes

1– Experience has shown certain Design Types have certain values

of X1 and X2 as function of Xbr as follows:

• Wound rotor : X1 = X2 = 0.5Xbr

• Design A : X1 = X2 = 0.5Xbr

• Design B : X1 = 0.4Xbr and X2 = 0.6Xbr

• Design C : X1 = 0.3Xbr and X2 = 0.7Xbr

• Design D : X1 = X2 = 0.5Xbr

2- For some design-class induction motors, a blocked-rotor test is

conducted under a test frequency, usually less than the normal

operating frequency so as to evaluate the rotor resistance

appropriately. If the test frequency is different from the rated

frequency, one can compute the total equivalent reactance at the

normal operating frequency (Xbr) as follows since the reactance is

directly proportional to the frequency:

'

br

test

ratedbr X

f

fX

Where X´br is blocked-rotor reactance at the test frequency

7

Example : 208 V, 60 Hz, six-pole Y-connected 25-hp design class B

induction motor is tested in the laboratory, with the following results:

DC test : 13.5 V , 64 A

No load test : 208 V , 22.0 A , 1200 W , 60 Hz

Blocked-rotor test : 24.6 V , 64.5 A , 2200 W , 15 Hz

Find R1 , R2 , X1 , X2 , and Xm of this motor ?

:Solution

: DC test From

R dc = V dc / I dc = 13.5 V/ 64 A = 0.2109 Ω

For Y-connected induction motor R1 = R dc / 2 = 0.2109 / 2 = 0.1054 Ω

:test Load No From

WPVVAI nlnlnl 1200,1203

208,22 11

2

2

1113

nl

nlnlnl

PIVQ

8

5.26093

120022120

2

2

1

nlQ

nl

nl

nlQ

VX

1

2

1

mnl XXX 1

2

5182.55.2609

120

rotor test :-blocked From

WPVVAI brbrbr 333.7333

2200,2.14

3

6.24,5.64 111

1762.05.64

333.73322

1

1

br

brbr

I

PR

0708.01054.01762.012 RRR br

2

1

2

111 brbrbrbr PIVQ

721.548)333.733(5.642.14 22

1 brQ

HzatI

QX

br

brbr 151318.0

5.64

721.54822

1

1'

5272.01318.015

60'

Hz

HzX

f

fX br

test

rated

br

For design B motor

X1 = 0.4 Xbr = 0.4 * 0.5272 = 0.2108 Ω

X2 = 0.6 Xbr = 0.6 * 0.5272 = 0.3163 Ω

Xm = Xnl – X1 = 5.5182 – 0.2108 = 5.3074 Ω So ,

9

HOME WORK

Q1: Test carried out on 3phase, Y-connected, 5HP ,208V ,4Pole

SCIM , with the following results:

DC test : 20 V , 25 A

No load test : 208 V , 4 A , 250 W , 60 Hz

Blocked-rotor test : 35 V , 12 A , 450 W , 15 Hz

Find R1 , R2 , X1 , X2 , and Xm of this motor ?

Q2: Test carried out on 3phase, Δ-connected, 30kW ,415V,

50 Hz SCIM , with the following results:

No load test : 415 V , 21 A , 1250 W

Blocked-rotor test : 100 V , 45 A , 2730 W

Find X1 , X2 , and Xm of this motor ?

1

Loading , braking & application of a 3phase induction motors

Introduction ;

• When selecting a 3-phase induction motor for an application several

motor types can fill the need , manufactures often specify the motor class

best suited to drive the load .

• 3-phase induction motors under 500 hp are standardized ,i.e. the

frames have standard dimensions. So, A 25 hp, 1725 rpm, 60 Hz motor

from one manufacture can be replaced by that of any other manufacturer

without having to change the mounting holes, shaft height, or the type of

coupling

• a motor must satisfy minimum requirements for starting torque, locked-

rotor current, overload capacity, and temperature rise.

Motor characteristics under load • High-inertia load stains a motor by prolonging the starting period.

– the starting currents in both the rotor and stator are high during starting.

– overheating from I²R losses becomes a problems.

– prolonged starting of very large motors will overload the utility

transmission network.

• Induction motors are often started on reduced voltage to:-

– limits the current drawn by the motor

– reduces the heating rate of the rotor

• Heat dissipated in the rotor during starting, from zero speed to rated

speed, is equal to the final kinetic energy stored in all the revolving

components .

LOAD CONSIDERATIONS There are three basic load types, and these types are classified by the

relationship of horsepower and speed.

1- CONSTANT TORQUE LOAD.

2- CONSTANT HORSE POWER LOAD.

3- VARIABLE TORQUE LOAD.

2

Constant torque applications are those that have the same torque at all

operating speeds, and horsepower varies directly with the speed. About

90% of all applications, other than pumps, are constant torque loads.

Examples of this type include conveyors, hoisting loads, surface winding

machines, positive displacement pumps and piston and screw

compressors.

Constant horsepower applications have higher values of torque at lower

speeds, and lower values of torque at higher speeds. Examples include

lathes, milling machines, and drill presses .

The drills in the diagram below are an example of a constant horsepower

application. When a larger hole is being drilled, the drill is operating at

low speed, but it requires a very high torque to turn the large drill in the

material. When a small hole is being drilled, the drill is operated at a high

3

speed, but it requires a very low torque to run the small drill in the

material.

The last basic load type is variable torque. The torque required varies as

the square of its speed, and horsepower requirements increase as the cube

of the speed. Examples include centrifugal pumps, turbine pumps,

centrifugal blowers, fans and centrifugal compressors.

Motor Classifications according to speed – For a given output power, a high-speed motor compared with a low-speed motor

• costs less

• is smaller sized

• has higher efficiency and power factor

–Low speed applications are often best served by using a high speed- motor

and a gear-box which is often required to modify operating speed especially

for very high-speed applications (> 3600 rpm).

4

Braking of a 3 phase induction motors:

1- Plugging : • In some applications, the motor and its load must come to a quick stop

– such braking action can be accomplished by interchanging two

stator leads ,the lead switching causes the revolving field to turn in

the opposite direction. Kinetic energy is absorbed from the

mechanical load causing the speed to fall.The absorbed energy is

dissipated as heat in the rotor circuit.

• plugging produces I²R losses that exceed the locked rotor losses.

• High rotor temperatures will result that may cause the rotor bars to

overheat or even melt. The heat dissipated in the rotor during plugging,

from initial speed to zero, is three times the original kinetic energy of all

revolving parts.

2- DC Braking: • An induction motor with high-inertial load can also come to a quick

stop by circulating dc current in the stator winding.

– any two stator terminals can be connected to a dc source.

– the dc current produces stationary N,S poles in the stator.

• the number of stationary poles are the same at the number of rotating

poles normally produced with ac currents.

– as the rotor bars sweeps past the dc field, an ac rotor voltage is induced

• the I²R losses produced in the rotor circuit come at the expense of the

kinetic energy stored in the revolving components.

– the motor comes to a rest by dissipating as heat all the kinetic energy.

• The benefit of dc braking is the far less heat that is produced.

– dissipated rotor losses is equal to the kinetic energy of the revolving

parts.

– energy dissipation is independent of the dc current magnitude

– the braking torque is proportional to the square of the dc braking

current.

Enclosure considerations The enclosure of the motor must protect the windings, bearings, and other

mechanical parts from moisture, chemicals,mechanical damage and

abrasion from grit. NEMA standards define more than 20 types of

enclosures under the categories of open machines, totally enclosed

machines, and machines with encapsulated or sealed windings.

The most commonly used motor enclosures are open dripproof, totally

enclosed fan cooled and explosionproof.

5

Open Dripproof(ODP):

The open dripproof motor has a free exchange of air with the ambient.

Drops of liquid or solid particles do not interfere with the operation at any

angle from 0 to 15 degrees downward from the vertical. The openings are

intake and exhaust ports to accommodate interchange of air. The open

dripproof motor is designed for indoor use where the air is fairly clean

and where there is little danger of splashing liquid.

ODP MOTOR

Totally Enclosed Fan Cooled (TEFC):

This type of enclosure prevents the free exchange of air between the

inside and outside of the frame, but does not make the frame completely

airtight. A fan is attached to the shaft and pushes air over the frame

during its operation to help in the cooling process. The ribbed frame is

designed to increase the surface area for cooling purposes. There is

also a totally enclosed non-ventilated (TENV) design which does not use

a fan, but is used in situations where air is being blown over the motor

shell for cooling, such as in a propeller fan application.

The TEFC style enclosure is the most versatile of all. It is used on pumps,

fans, compressors, general industrial belt drive and direct connected

equipment. Special protection can be added to the TEFC motor to help it

withstand hostile environments such as chemical and pulp and paper

applications.

TEFC MOTOR

6

Explosion proof

The explosion proof motor is a totally enclosed machine and is designed

to withstand an explosion of specified gas or vapor inside the motor

casing and prevent the ignition outside the motor by sparks, flashing

or explosion. These motors are designed for specific hazardous purposes,

such as atmospheres containing gases or hazardous dusts. For safe

operation, the maximum motor operating temperature must be below the

ignition temperature of surrounding gases or vapors. Explosion proof

motors are designed, manufactured and tested under the rigid

requirements of the Underwriters Laboratories.

ENVIRONMENTAL CONSIDERATIONS : AC motors that are properly selected and used should give many years of

satisfactory service. Motor life is prolonged by keeping the motor cool,

dry, clean and lubricated. Choosing the best motor for the application and

the environment will insure that long life. Because so many conditions

contribute toward short service life, it is impractical to set forth all

possibilities, but some discussion will help point out the need for

application analysis where trouble is experienced.

Overheating:

Heat is one of the most destructive stresses causing premature motor

failure. Overheating occurs because of motor overloading, low or

unbalanced voltage at the motor terminals, excessive ambient

temperatures, or poor cooling caused by dirt or lack of ventilation. If the

heat is not dissipated, insulation failure and possibly lubrication and

bearing failure can damage a motor.

Ambient Temperature:

For the purposes of standardization, a value of 40°C has been selected as

the standard industrial ambient. This value covers a large range of

practical ambient temperatures encountered throughout the world.

Temperature rise:

This aspect is related to the maximum allowable temperature increase

above the defined ambient that the winding cannot exceed when the

motor is operating at the name plate or rated power output.

7

Rise by Resistance : One method of determining winding temperature rise (Δt)during motor testing is to measure the electrical resistance between the leads of each phase of the winding. Each phase consists of a set of windings (coils) made from a conductor (typically copper) that has a resistance that is measured in ohms. This value is measured and recorded prior to initiating a load test to determine temperature rise, along with the actual ambient temperature, The test is performed by applying a load equal to the designed power capability of the motor by dynamometer or other means . Under test conditions, the winding temperature of a continuous duty motor must achieve steady-state stabilization while operating under the defined power output or loading value. The heating developed under loading results in a resistance change of the winding conductors.

etemperaturroomtesttwhere

R

tRRt

cold

coldhot

1

1 235*

Service Factor :NEMA defines maximum allowable temperature rise for power output conditions greater than the nominal rated conditions. The commonly defined service factor is 1.15, but other multipliers can be applied provided the maximum temperature rise is limited to that specified for the service factor condition. A motor with a 1.15 service factor is expected to deliver 15 percent more power than the normal full load condition. Moisture: Moisture should be kept from entering a motor. Water from

splashing or condensation seriously degrades an insulation system. The

water alone is conducting. The proper type of motor should be chosen for

use in a damp environment.

Contamination: No conducting contaminants such as factory dust and

sand gradually promote over-temperature by restricting cooling air

circulation. In addition, these may erode the insulation and the varnish,

gradually reducing their effectiveness.

Altitude: Standard motor ratings are based on operation at any altitude up

to 3300 feet (1000 meters). High altitude derating is required above 3300

feet because of lower air density.Less dense air reduces the ability for

heat to be dissipated into the air from the surfaces of equipment With

reduced thermal transfer ability, a motor cannot effectively dissipate heat

losses, which require adjustments in the temperature rating, and

corresponding power capability of a given motor as defined in Table

below.

8

Induction motor operating as a generator : When run faster than its synchronous speed , an induction motor runs as a

generator called induction generator . It convertsthe mechanical energy it

receives into electrical energy and this energy is released by by the stator .

As soon as motor speed exceeds its synchronous speed , it starts delivering

active power ( P) to the three phase line . However , for creating its own

magnetic field ,it absorbs reactive power (Q)from the line to which it is

connected.The reactive power can also be supplied by a group of capacitors

connected across its terminals.If capacitance is insufficent , the generator

voltage will not build up .Induction generator is widely used today in wind

turbines ,which convert wind energy into electrical energy.

Complete torque -speed curve of a three phase induction machine

1

Introduction to Synchronous Machines

Definition:

A synchronous machine is an ac rotating machine whose speed under

steady state condition is proportional to the frequency of the current in its

armature. The magnetic field created by the stator currents rotates at the

synchronous speed ,and that created by the field current on the rotor is

rotating at the synchronous speed also, and a steady torque results. So,

these machines are called synchronous machines because they

operate at constant speeds and constant frequencies under steady-

state conditions. Synchronous machines are commonly used as

generators especially for large power systems, such as turbine generators

and hydroelectric generators in the grid power supply. Because the rotor

speed is equal to the synchronous speed of stator magnetic field,

synchronous motors can be used in situations where constant speed drive is

required. Since the reactive power generated by a synchronous machine can

be adjusted by controlling the magnitude of the rotor field current,

unloaded synchronous machines are also often installed in power systems

for power factor correction or for control of reactive kVA flow. Such

machines, known as synchronous condensers, and may be more

economical in the large sizes than static capacitors.

The bulk of electric power for everyday use is produced by polyphase

synchronous generators( alternators), which are the largest single-unit

electric machines in production. For instance, synchronous generators

with power ratings of several hundred megavolt-amperes (MVA) are

fairly common, and it is expected that machines of several thousand

megavolt- amperes will be in use in the near future. Like most rotating

machines, synchronous machines are capable of operating both as a

motor and as a generator. They are used as motors in constant-speed

drives, and where a variable-speed drive is required ,a synchronous

motor is used with an appropriate frequency changer such as an

inverter. As generators, several synchronous machines often operate

in parallel, as in a power station. While operating in parallel, the

generators share the load with each other; at a given time one of the

generators may not carry any load. In such a case, instead of shutting

down the generator, it is allowed to "float" on the line as a

synchronous motor on no-load.

2

Construction of synchronous machines : The synchronous machine has 3 phase winding on the stator and a d.c. field winding

on the rotor.

1. Stator : It is the stationary part of the machine and is built up of sheet-steel laminations

having slots on its inner periphery. A 3-phase winding is placed in these slots. The

armature winding is always connected in star and the neutral is connected to ground.

2. Rotor : The rotor carries a field winding which is supplied with direct current through two slip

rings by a separate d.c. source. Rotor construction is of two types, namely;

(i) Salient (or projecting) pole type .

(ii) Non-salient (or cylindrical) pole type .

(i) Salient pole type:

In this type, salient or projecting poles are mounted on a large circular steel frame

which is fixed to the shaft of the alternator as shown in Fig. (1). The individual field

pole windings are connected in series in such a way that when the field winding is

energized by the d.c. exciter, adjacent poles have opposite polarities.

Low-speed alternators (120 - 400 r.p.m.) such as those driven by water turbines have

salient pole type rotors due to the following reasons:

(a) The salient field poles would cause .an excessive windage loss if driven at high

speed and would tend to produce noise.

(b) Salient-pole construction cannot be made strong enough to withstand the

mechanical stresses to which they may be subjected at higher speeds.

Since a frequency of 50 Hz is required, we must use a large number of poles on the

rotor of slow-speed alternators. Low-speed rotors always possess a large diameter

to provide the necessary space for the poles. Consequently, salient-pole type rotors

have large diameters and short axial lengths.

Fig. 1 salient pole rotor

3

(ii) Non-salient pole(cylindrical) type :

In this type, the rotor is made of smooth solid forged-steel radial cylinder having a

number of slots along the outer periphery. The field windings are embedded in

these slots and are connected in series to the slip rings through which they are

energized by the d.c. exciter. The regions forming the poles are usually left

unslotted as shown in Fig. (2). It is clear that the poles formed are non-salient i.e.,

they do not project out from the rotor surface.

Fig. 2 cylindrical rotor

High-speed alternators (1500 or 3000 r.p.m.) are driven by steam turbines and

use non-salient type rotors due to the following reasons:

(a) This type of construction has mechanical robustness and gives noiseless

operation at high speeds.

(b) The flux distribution around the periphery is nearly a sine wave and hence

a better e.m.f. waveform is obtained than in the case of salient-pole type.

Since steam turbines run at high speed and a frequency of 50 Hz is required, we

need a small number of poles on the rotor of high-speed alternators (also called

turboalternators). We can use not less than 2 poles and this fixes the highest

possible speed. For a frequency of 50 Hz, it is 3000 r.p.m. The next lower speed is

1500 r.p.m. for a 4-pole machine. Consequently, turboalternators possess 2 or 4

poles and have small diameters and very long axial lengths.

4

Classification of synchronous machines according to the form of excitation:

1. Brushes excitation systems:

The field structure is usually the rotating member of a synchronous machine

and is supplied with a dc-excited winding to produce the magnetic flux. This dc

excitation may be provided by a self-excited dc generator mounted on the same

shaft as the rotor of the synchronous machine. This dc generator is known as

exciter. The direct current generated inside exciter is fed to the synchronous

machine field winding. In slow-speed machines with large ratings, such as

hydroelectric generators, the exciter may not be self-excited. Instead, a pilot

exciter, which may be self-excited or may have a permanent magnet, activates the

main exciter.

Fig 1 Block Diagram of Excitation system by pilot &main

Fig. 1 Brushes Excitation system for a synchronous machine

2.Brushless systems:

This type of excitation has a shaft-mounted bridge rectifier, that rotate with the

rotor, thus avoiding the need for brushes and slip rings.

Fig.2 Brushless Excitation system for a synchronous machine

5

3.static systems:

This type of exctation is widely used in small size alternators, in which portion of

the AC from each phase of synchronous generator is fed back to the field winding

as a DC excitation through a system of transformer ,rectifiers and reactors.

External source of a DC is necessary for initial excitation of the field windings.

Advantages of stationary armature ;

The field winding of an alternator is placed on the rotor and is connected to d.c.

supply through two slip rings. The 3-phase armature winding is placed on the stator.

This arrangement has the following advantages:

(i) It is easier to insulate stationary winding for high voltages , because they are not

subjected to centrifugal forces .

(ii) The stationary 3-phase armature can be directly connected to load without going

through large, unreliable slip rings and brushes.

(iii) Only two slip rings are required for d.c. supply to the field winding on the rotor.

Since the exciting current is small, the slip rings and brush gear required are of light

construction.

(iv) Due to simple and robust construction of the rotor, higher speed of rotating d.c.

field is possible. This increases the output obtainable from a machine of given

dimensions.

Cooling :

Because synchronous machines are often built in extremely large sizes, they

are designed to carry very large currents. A typical armature current density

may be of the order of 10 A/mm2 in a well-designed machine. Also, the

magnetic loading of the core is such that it reaches saturation in many regions.

The severe electric and magnetic loadings in a synchronous machine produce

heat that must be appropriately dissipated. Thus the manner in which the

active parts of a machine are cooled determines its overall physical structures.

In addition to air, some of the coolants used in synchronous machines include

water, hydrogen, and helium.

6

Damper Bars :

So far we have mentioned only two electrical windings of a synchronous machine:

the three-phase armature winding and the field winding. We also pointed out that,

under steady state, the machine runs at a constant speed, that is, at synchronous

speed. However, like other electric machines, a synchronous machine undergoes

transients during starting and abnormal conditions. During transients, the rotor may

undergo mechanical oscillations and its speed deviates from the synchronous speed,

which is an undesirable phenomenon. To overcome this, an additional set of

windings, resembling the cage of an induction motor, is mounted on the rotor. When

the rotor speed is different from the synchronous speed, currents are induced in

the damper windings .the damper winding acts like the cage rotor of an induction

motor, producing a torque to restore the synchronous speed. Also, the damper bars

provide a means of starting to the synchronous motors, which is otherwise not self-

starting. Fig6 shows the damper bars on a salient rotors

Fig .6 salient pole rotor showing the field winding and damper bars

7

Differences between three phase induction machine and synchronous machine

Synchronous machine Three phase induction machine

Stator phase are star connected only

connected Stator phases either star or delta

Rotor windings are fed by dc source.

Rotor windings are not fed by electricity,

currents flow through rotor due to induction

process.

Run at synchronous speed for both motor and

generator

Run below synchronous speed, as a motor

Run above synchronous speed, as a generator

Need damper bars to start , as a motor

Self starting , as a motor

Operate with lagging, leading, and unity power

factor

Operate with lagging power factor only

Complex in construction , expensive.

Simple in construction , rugged, low

maintenance, cheap , especially in squirrel cage

type.

Active in both low and high speed operation

, as a motor

Not active in low speed operation, as a motor

Used widely in generation of electricity

Not active in generating mode

Precise speed regulation, as a motor

Difficult in speed regulation, as a motor

8

THREE PHASE STATOR WINDINGS

The stator windings for alternating-current motors and generators are alike. It should be

noted that direct-current and alternating-current windings differ essentially by the former

being of the closed-circuit type (through commutator) ,while alternating-current windings

are of the open-circuit type.

Types of A-C Windings: With reference to the arrangements of coils used in three phase stator, windings may be

divided into two general classes as follows:

I. Distributed Windings:

1. Spiral or chain.

2. Lap.

3. Wave.

II. Concentrated Windings:

1. Lap.

2. Wave.

Distributed Windings: An armature winding which has its conductors of any one phase under a single pole placed

in several slots, is said to be distributed. When these conductors are bunched together in

one slot per pole per phase, the winding is called concentrated. It is usual in a distributed

winding to distribute the series conductors in any phase of the winding among two or more

slots under each pole. A distributed winding has two principal advantages, first, a distributed

winding generates a voltage wave that is nearly a sine curve, secondly, copper is evenly

distributed on the armature surface, therefore, heating is more uniform and this type of

winding is more easily cooled.

Concentrated Windings: The concentrated winding gives the largest possible emf from a given number of conductors

in the winding. That is for a definite fixed speed and field strength in an alternator, the

concentrated winding requires a less number of conductors than a distributed winding, but

increases the number of turns per coil.

Lap and Wave Windings:

Both distributed and concentrated windings make use of lap and wave connections. These

arrangements are in principle the same as used in direct-current windings. The diagrams of

Figs. 1 and 2 show distributed lap and wave windings .

9

Fig. 1 Single-phase lap winding

Fig 2. Three-phase wave winding , using one slot per pole per phase, star-connected.

Chain Winding : In this winding as shown in Fig. 3 there is only one coil side in a slot. An odd or even

number of conductors per slot may be used but several shapes of coils are required since the

coils enclose each other. The number of coils required in this winding is also small

compared with other windings. This type of winding is mainly used in alternating current

generators.

Fig. 3 Three-phase chain winding

10

Single layer and double layer winding :

In double layer winding two coil sides are located in each armature slot [Fig. 4 ].

If there is only one coil side located in each armature slot, the winding is called

single layer.

Fig. 4 A double layer winding. One side of the coil lies in the top of the slot and the other side in the

bottom of the of the slot.

Full pitch and fractional pitch windings:

For a three-phase winding the total number of coil sides or the total number of slots should

be just divisible by three (the number of phases) and sometimes by the number of poles.

This will result in a full pitch winding, that is, a winding in which a coil spans exactly the

distance between the centers of adjacent poles. If the coil spans less than this distance, so

that its two sides are not exactly under the centers of adjacent poles at the same time, it is

said to have a fractional-pitch. When a fractional pitch is used, the total number of slots

per phase must be a whole number. The fractional-pitch coils are frequently used in a.c.

machines for two main reasons. First, less copper is required per coil and secondly the

waveform of the generated voltage is improved .The number of stator slots, divided by the

number of poles gives a value of the pole pitch expressed in terms of the slots. Coil pitch

is expressed as a fraction of the pole pitch, in slots, or in electrical degrees. In the case of a

6 pole machine having 72 stator slots, and a double-layer winding, the pole pitch would be

12 slots(72 / 6). If the coil pitch were given as 2/3, this would be 120° ( 2/3 * 180°) or 8

slots (2/3 * 12). A full coil pitch for this winding would be 180 degrees, or 12 slots. A full

pitch winding is one in which the coil pitch is equal to the pole pitch, and a fractional pitch

winding is one in which the coil pitch is not equal to the pole pitch. For the coils used in

small machines round insulated wire is most employed. These coils are either wound in

the slots by hand or assembled by use of specially formed coils wound in forms and

insulated before being placed in the slots. Such formed coils are usually used except in

cases where the slots are closed or nearly closed. For large machines where the amperes to

be carried in each armature circuit is a large value, copper straps are frequently employed

for making up the armature coils. In very large machines a copper bar is used instead of

the copper straps. In such a case one bar serves as a conductor of a coil having one turn per

slot. The copper bars are connected to the end connections of the coils by brazing, welding

or bolting. In all cases, whatever the construction of the coil used, the slots must be

properly insulated with mica, polyester film or other suitable insulating material according

11

to the adequate class of insulation. The coil throw refers to the start of coil from one stator

slot to the finish of this coil at another stator slot . Figures 5 (a) , (b) , (c), and (d) show

four different coils, in each of them the pole pitch is of 12 slots while the coil pitch in (a) is

11 slots , in (b) is 13 slots, in (c) is 9 slots, and in (d) is 15 slots .

Fig. 5. Possible pitch for one coil per slot windings(pole pitch=24 slots/2poles=12) (a) coil throw 1 to 12,

coil pitch=12-1=11 (b) coil throw 1 to 14, coil pitch=14-1=13 (c) coil throw 1 to 10, coil pitch=10-1=9 ,

(d) coil throw 1 to 16, coil pitch=16-1=15 .

Phase belt and phase spread :

A group of adjacent slots belonging to one phase under one pole pair is known

as phase belt. The angle subtended by a phase belt is known as phase spread.

The 3-phase windings are always designed for 60° phase spread . Fig. (6)

shows a 2-pole, 3-phase double-layer, full pitch, distributed winding for the

stator of an alternator. There are 12 slots and each slot contains two coil sides.

The coil sides that are placed in adjacent slots belong to the same phase such

as( a1, a3 ) or (a2, a4 )constitute a phase belt. Since the winding has double-

layer arrangement, one side of a coil, such as (a1), is placed at the bottom of a

slot and the other side (- a1 ) is placed at the top of another slot spaced one pole

pitch apart. Note that each coil has a span of a full pole pitch or 180 electrical

degrees. Therefore. the winding is a full-pitch winding. Note that there are 12

total coils and each phase has four coils. The four coils in each phase are

connected in series so that their voltages aid. The three phases then may be

connected to form Y or -connection. Fig. (7) shows how the coils are

12

connected to form a Y-connection. Any star diagram can be readily changed

into a corresponding delta diagram by opening the star points and connecting

the inner end of phase A to the outer end of phase B, the inner end of phase B

to the outer end of phase C , and the inner end of phase C to the outer end of

phase A .

Fig. 6 Fig. 7

Simple rule for checking proper phase relationship in a three-phase winding:

A fundamental consideration when checking the instantaneous flow of current

in a three phase circuit, is to imagine that when the current flows in the same

direction in two legs of the circuit, it flows in the opposite direction in the third

leg. This principle can be applied to both motors and generators. In Figure 8, it

must be supposed that current flows in all three leads of the star connection

toward the point of the star connection. And that in the case of a delta

connection the current flows around the three sides of the delta in the same

direction. Then in either case for a three-phase winding, the polarity of each of

the pole-phase groups will alternate regularly around the winding and can be

indicated by arrows as in Fig. 8. By the use of this scheme there is no chance

13

for a reversal of a phase to be passed by not noticed when checking the

winding.

. Fig. 8 Simple scheme of alternately reversing arrows of pole-phase groups to check correct phase

polarity of a 3-phase winding. It is supposed in this case that current flows in the three leads

toward the star points of the winding which are indicated thus ( *) .

Why the alternators have their windings connected in star?

Because the star connection has the following advantages over delta connection:

(i) To obtain a desired voltage between outside terminals of a generator, the voltage built up in any one phase need be only 58 percent(1 / √3) of the terminal value. Hence only 58 percent of the turns required for a Δ-connected armature are necessary with a consequent lowering of insulation cost.

(ii) A star connected winding offers the advantage of a fourth or neutral lead making possible the advantages of a four -wire system, with or without grounded neutral.

(iii)The wave shape of a star-connected winding is improved, owing to the elimination of third harmonics and multiples of third harmonics from the terminal voltage. Fig.9 shows a star connected armature( stator), The terminal voltages E12, E23 and,E31 are 120 electrical degrees apart. In the third harmonics the e.m.f.'s of three phases (e01 ,e02 ,e03) are being in phase(3 X 120º = 360º= 0º ),to obtain terminal voltages, the resultant is zero. Hence no third harmonics appears between terminals.The circulating currents from the third harmonics cause unnecessary losses and dangerous heating in Δ-connected alternator, where it is not in the Y-connected alternator case. In addition the use of

14

a(5/6)th pitch winding in three-phase Y-connected generators reduces the fifth and seventh harmonics, if pesent, to almost nil, so the lowest harmonics that can be present is eleventh.

Fig. 9

15

Windings factors:

The stator winding of synchronous machine is distributed over the entire stator.

The distributed winding produces nearly a sine waveform and the heating is

more uniform. Likewise, the coils of armature winding are not full-pitched i.e.,

the two sides of a coil are not at corresponding points under adjacent poles. The

fractional pitched armature winding requires less copper per coil and at the

same time waveform of output voltage is improved. The distribution and

pitching of the coils affect the voltages induced in the coils. We shall discuss

two winding factors:

(i) Distribution factor (Kd).

(ii) Pitch factor (Kp).

(i) Distribution factor (Kd) :

A winding with only one slot per pole per phase is called a concentrated

winding. In this type of winding, the e.m.f. generated/phase is equal to the

arithmetic sum of the individual coil e.m.f.s in that phase. However, if the

coils/phase are distributed over several slots in space (distributed winding), the

e.m.f.s in the coils are not in phase (i.e., phase difference is not zero) but are

displaced from each by the slot angle (The angular displacement in electrical

agrees between the adjacent slots is called slot angle). The e.m.f./phase will be

the phasor sum of coil e.m.f.s. The distribution factor Kd is defined as:

The distribution factor can be determined by constructing a phasor diagram for

the coil e.m.f.s. Let m = 3. The three coil e.m.f.s are shown as phasors AB, BC

and CD [See Fig. 10 (i)] each of which is a chord of circle with centre at O and

subtends an angle at O. The phasor sum of the coil e.m.f.s subtends an angle

16

m(Here m = 3) at O. Draw perpendicular bisectors of each chord such as Ox,

Oy etc [See Fig. 10 (ii)].

Note that ( m is the phase spread.

Fig. 10

(ii) Pitch factor (Kp) : A coil whose sides are separated by one pole pitch (i.e., coil span is 180electrical) is

called a full-pitch coil. With a full-pitch coil, the e.m.f.s induced in the two coil sides a in

phase with each other and the resultant e.m.f. is the arithmetic sum of individual e.m.fs.

However the waveform of the resultant e.m.f. can be improved by making the coil pitch

less than a pole pitch. Such a coil is called short-pitch coil. This practice is only possible

with double-layer type of winding The e.m.f. induced in a short-pitch coil is less than that

of a fullpitch coil. The factor by which e.m.f. per coil is reduced is called pitch factor Kp. It

is defined as:

Consider a coil AB which is short-pitch by an angle Өelectrical degrees as shown in Fig.

(11). The e.m.f.s generated in the coil sides A and B differ in phase by an angle Өand can

17

be represented by phasors EA and EB respectively as shown in Fig. (12). The diagonal of the

parallelogram represents the resultant e.m.f. ER of the coil.

Fig. 11 Fig. 12

For a full-pitch winding, Kp = 1. However, for a short-pitch winding, Kp < 1.

Note that Өis always an integer multiple of the slot angle .

18

SYNCHRONOUS GENERATOR

The machine which produces 3-phase electrical power from mechanical power is called

an alternator or synchronous generator. Alternators are the primary source of all the

electrical energy we consume. These machines are the largest energy converters found

in the world.

Alternator Operation : Like the dc generator, a synchronous generator functions on the basis of Faraday's

law, which state if the flux linking the coil changes in time, a voltage is induced

in a coil. Stated in another form, a voltage is induced in a conductor if it cuts

magnetic flux lines. The rotor winding is energized from the d.c. exciter and alternate N

and S poles are developed on the rotor. When the rotor is rotated by a prime mover, the

stator or armature conductors are cut by the magnetic flux of rotor poles. Consequently,

e.m.f. is induced in the armature conductors due to electromagnetic induction. The

induced e.m.f. is alternating since N and S poles of rotor alternately pass the armature

conductors. The frequency of induced e.m.f. is given by;

Where N = speed of rotor in r.p.m. ,

P = number of rotor poles.

The magnitude of the voltage induced in each phase depends upon the rotor flux, the

number and position of the conductors in the phase and the speed of the rotor.

[Fig. 1 (i)] shows star-connected armature winding and d.c. field winding. When the rotor

is rotated, a 3-phase voltage is induced in the armature winding.. The magnitude of e.m.f.

in each phase of the armature winding is the same. However, they differ in phase by 120°

electrical as shown in the phasor diagram [ Fig. 1 (ii)].

Fig. 1

19

Considering the round rotor synchronous generator in Fig.2,which shows a simple

consentrated stator winding ( 1 slot / pole / phase).

Fig .2 (a) A round rotor synchronous generator

(b) Air gap flux density distribution produced by the rotor excitation

and assuming that the flux density in the air gap is uniform implies that sinu-

soidally varying voltages will be induced in the three coils aa' ,bb' and cc' if the rotor,

rotates at a synchronous speed, Ns. Also if Φр is the flux per pole, ω is the angular

frequency, and T is the number of turns in phase a (coil aa'), then the voltage

induced in phase a is given as:

ea = ω T Φр sin ωt

= Em sin ωt

Where Em = ω T Φр , and ω =2 π f .

Because phases b and c are displaced from phase a by +_ 120º , the corresponding

voltages may be written as

eb = Em sin (ωt – 120º)

ec = Em sin (ωt + 120º)

These voltages are sketched in Fig. 3 and correspond to the voltages from a three-

phase generator.

20

Fig. 3 A three-phase voltage produced by a three-phase synchronous generator

Generated E.M.F. Equations:

The magnetic flux cut by one armature conductor, when the rotor of an alternator is made to revolve through one revolution, is Φр * P, where Φр is the magnetic flux per pole and P is the number of poles. If N is the speed in revolution per minute, then the flux cut per second becomes

Φр * P *( N /60)

Since 1 volt is generated when 1 weber of flux is cut per second, the average voltage generated in this conductor becomes

Eav = Φр * P *( N /60) volts

If the total number of conductors on the armature is Z and they are connected into A parallel paths, the average voltage between terminals becomes Eav per conductor * ( Z / A )

In case of alternating current generators the e.m.f. depends not only upon the total flux cut per second but also upon the way in which the flux and the conductors are distributed. A change in distribution of flux changes the relative values of the maximum and effective (r.m.s.) e.m f.'s.

In addition, the e.m.f. built up in any one conductor, when considered vectorially, cannot always be added directly to that of another as there may be a phase displacement between them. The instantaneous values, though can be added algebraically, but in adding the effective values it is necessary to consider the phase differences between the different e.m.f.'s to be added. In order to take these factors in account, the flux distribution and winding types must be known.

Fig.4(a) shows the space distribution of the airgap flux of an alternator. In this case, the flux density B is assumed to be sinusoidal in space when measured around the inside periphery of the stator. This flux density B can be expressed as

B=Bmax sinθ.

where θ is measured from the position midway between the poles ; B is the flux density measured in webers per length of the field pole (L ) as shown in Fig.4(b) per length of pole pitch arc ; and Bmax is the maximum flux density produced by a pole.

21

The total flux per pole is

Φp =ο∫π LBmax sin θ dθ = -| LBmax cos θ |ο

π =2 LBmax Weber ...(i)

As this flux wave is moved around the air gap the conductors a and b of the stator coil will have voltages induced in them.

Fig 4 Shapes of flux density waves

The voltage induced in conductor a will be

ea=BL v ...(ii)

where v is the velocity of the flux wave in radians per second, or v = 2 π f , f being the frequency of the flux wave or e.m.f. induced.

The voltage in conductor a is

ea=BmaxL 2π f sin θ

But θ = ω t where ω is the angular velocity of the rotor in radians per second and t is the time of rotation from the position shown in Fig.4(a). Therefore,

ea = BmaxL 2π f sin ω t ...(iii)

Substituting equation (i) in equation (iii)

ea = (Φp /2 ) *2π f sin ω t ...(iv)

Likewise, the voltage induced in conductor b

eb = (Φp /2 ) *2π f sin ω t ...(v)

22

and the voltage at the terminals of the turn made up of the conductors a and b is

eturn = Φp 2π f sin ω t ...(vi)

If the single turn on the armature (stator) were replaced by a coil of T turns, the voltage per coil would be

eeoil = Φp 2π f T sin ω t ...(vii)

The effective value of generated e.m.f.

Er.m.s = Emax / √2 = (2π / √2) f Φp T volts (r.m.s) ...(viii)

Er.m.s =4.44 f Φp T volts (r.m.s) ...(ix)

As the r.m.s. value is related to the average value so,

Form factor (kf) = (r.m.s. value) / (average value)

* For sinusoidal wave of e.m.f., kf = 1.11

Er.m.s . = 4 kf T f Φp volts

Also, since the induced e.m.f.'s in the conductors of an alternator are not all in phase, the above relation for Er.m.s. must be multiplied by kp and kd, the pitch factor and the distribution (breadth) factor.

Therefore,

Er.m.s. =4 kf kp k d f T Φp volts per phase ...(x)

* For full pitched and concentrated windings kd =1 and kP =1.

If the alternator is star connected, and neglecting the effect of armature reaction, the alternator line voltage is,

EL = √ 3 (Er.m.s. per phase) , so

EL = √ 3( 4 kf kp kd f T Φp) volts.

23

Synchronous Generator Characteristics :

( i ) Magnetisation curve:

A plot of the exciting current versus terminal voltage of alternator is known as

the magnetisation curve. This magnetisation curve is obtained by passing different

values of currents in exciting windings, thereby giving, correspondingly

different values of terminal voltage. The no load magnetisation curve is

shown in Fig.5-I] ,which has the same general shape as B-H curve of armature

steel.

The full load magnetisation characteristics with unity power factor and with 0.8 lagging power factor have been shown at ( II ) and ( III ) in Fig.5.

Fig. 5 Magnetization Curve

Fig. 5 Magnetization curves for alternator at different loading conditions

( ii ) Load characteristics :

If the speed and exciting current remain constant, the terminal voltage of the alternator changes with the load currents. The plot between the terminal voltage and the load (or armature) current of an alternator is known as load characteristics. An increase in the armature (or load) currents make the terminal voltage drops. This has been shown in Fig. 6. The drop in terminal voltage is attributed to many reasons but primarily because of the following:

( a ) Resistance and reactance of the armature (or stator) winding.

( b ) Armature reaction.

The resistance and reactance of the armature winding causes the drop in

generated e.m.f. (voltage) ,whereas the armature reaction weakens the magnetic

field and thereby decreases the generated e.m.f. (voltage).

24

The magnitude of the effect of armature reaction depends upon, the power factor

of load i.e. angle of lag or lead of the stator (armature) current. In case of a

unity power factor of load, each phase of alternator when connected to the

load takes a current which is in phase with its generated voltage. But the

magnetic field is strengthened, instead of weakening, if the load (or armature)

current, is leading. In the above case, when the power factor is leading, the drop

in voltage due to resistance and reactance of stator winding may be less than the

increase in voltage due to armature reaction. Thus the terminal voltage on

load may be more than that at no-load, if the angle of lead of the load (or

armature) current is sufficient.

Fig. 6. Load Characteristic of alternator

( iii ) Effect of variation of power factor on terminal voltage :

The load characteristics at different power factors with leading and lagging armature currents are shown in Fig. 7. If the load, current and excitation are kept constant, the terminal voltage falls on changing the power factor from leading to lagging one.This effect is because of the armature flux helping the main flux, in case p.f. is leading, hence generating more e.m,f. and the armature flux, opposing the main flux, in case the p.f. is lagging, hence generating less e.m.f., Therefore, the terminal voltage at lagging power factor decreases from that on leading p.f. because of decrease in generated e.m.f.

25

Fig. 7

Armature Leakage & Synchronous Reactances :

When the current pass through the stator conductors the flux is set up, and a portion of this flux does not cross the air gap but completes the path inside the stator as shown in Fig. 8. This flux is known as leakage flux,which sets up an e.m.f. in the stator winding .This e.m.f. leads the load current by 90° and proportional to the magnitude of load current. This e.m.f. is due to the leakage inductance of the armature winding. The magnitude of the leakage inductance in practical units,henrys, is given by the general equation :

L = (Flux in webers per amperes) * (No. turns) henrys

And leakage reactance per phase,

Xl = ω L = 2 π f L ohms/ph

Xl causes a voltage drop in alternator terminal voltage and this drop is equal to an e.m.f. set up by the leakage flux. Also, there is another source causes voltage drop , that is due to armature reaction which can be represented by a fictitious reactance Xa.

Xl + Xa = Xs

Where Xs is the per phase synchronous reactance of armature winding.

Fig 8

26

Performance Of A Round-Rotor Synchronous Generator :

At the outset we wish to point out that we will study the machine on a per phase basis,

implying a balanced operation. Thus let us consider a round-rotor machine operating as

a generator on no-load. Variation of the terminal voltage with the exciting current

(field current) is shown in Fig. 5-I , and is known as the open-circuit characteristic

of a synchronous generator. Let the open-circuit phase voltage be Eo for a certain

field current If. Here Eo is the internal voltage of the generator. We assume that If is

such that the machine is operating under unsaturated condition. Next we short-circuit

the armature at the terminals, keeping the field current unchanged (at If), and

measure the armature phase current Ia In this case, the entire internal voltage Eo is

dropped across the internal impedance of the machine. In mathematical terms,

Eo = IaZs

and Zs is known as the synchronous impedance , which is equal to

Zs= Ra + j XS

If the generator operates at a terminal voltage Vt, while supplying a load corresponding

to an armature current Ia, then

Eo = Vt + Ia (Ra +

j XS)

In an actual synchronous machine, except in very small ones, we almost always

have XS >> Ra , in which case Zs ≈ j XS. Among the steady-state characteristics of a

synchronous generator, its voltage regulation and power-angle characteristics are the

most important ones.

We define the voltage regulation of a synchronous generator at a given load as

percent voltage regulation = (Eo - Vt) / Vt * 100 %

where Vt is the terminal voltage on load and Eo is the no-load terminal voltage.

The voltage regulation is dependent on the power factor of the load. the voltage

regulation for a synchronous generator may even become negative. The angle

between Eo and Vt is defined as the power angle, δ. Notice that the power angle, δ, is

not the same as the power factor angle, φ. To justify this definition, we consider

Fig. 9, from which we obtain

Ia XS cos φ = Eo sin δ …(i)

27

Fig. 9 Phasor diagram of round rotor generator

Now, from the approximate equivalent circuit (assuming that XS >> Ra )as shown

in Fig 10-a,

the power delivered by the generator = power developed, Pd = Vt Ia cos φ ,

which follows from Fig. 9 also. Hence, in conjunction with the (i) equation, we get

Pd = (Eo Vt / XS ) sin δ per phase …(ii)

Pd = 3(Eo Vt / XS ) sin δ for three phases

Which shows that the internal power of the machine is porportional to sin δ,

Equation (ii) is often said to represent the power-angle characteristic of a round

rotor synchronous machine.

Fig. 10-b shows that for a negative δ, the machine will operate as a motor.

Fig 10 : (a) an approximate equivalent circuit of synchronous machine

(b) power-angle characteristics of a round-rotor synchronous machine

28

Performance Of A Salient-Pole Synchronous Generator :

Because of saliency, the reactance measured at the terminals of a salient-rotor machine

will vary as a function of the rotor position. This is not so in a round-rotor machine.

Thus a simple definition of the synchronous reactance for a salient-rotor machine is

not immediately forthcoming. To overcome this difficulty, we use the two-reaction

theory proposed by "Andre Blondel". The theory proposes to resolve the given

armature mmf's into two mutually perpendicular components, with one located along

the axis of the rotor salient pole, known as the direct (or d )axis and with the other in

quadrature and known as the quadrature (or q )axis. Correspondingly, we may

define the d-axis and q-axis synchronous reactances, Xd and Xq for a salient-pole

synchronous machine. Thus, for generator operation, we draw the phasor diagram of

Fig. 11. Notice that Ia has been resolved into its d- and q-axis (fictitious) components,

Id and Iq With the help of this phasor diagram, we obtain

Id = Ia sin ( δ + φ )

Iq = Ia cos ( δ + φ )

Vt sin δ = Iq Xq = Ia Xq cos ( δ + φ )

From these we get

Vt sin δ = Ia Xq cos δ cos φ - Ia Xq sin δ sin φ

Or (Vt + Ia Xq sin φ) sin δ = Ia Xq cos δ cos φ

Dividing both sides by cos δ and solving for tan δ yields

tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ) …(iii)

With δ known (in term of φ), the voltage regulation may be computed from

Eo = Vt cos δ + Id Xd

Percent regulation = (Eo - Vt) / Vt *100%

Fig. 9 Phasor Diagram of a Salient-Pole Generator

Fig 11 Phasor diagram of salient-pole generator

29

In fact, the phasor diagram depicts the complete performance characteristics of the

machine.

Let us now use Fig. 11 to drive the power-angle characteristics of a salient-pole

generator. If armature resistance is neglected, Pd = Vt Ia cos φ, Now, from Fig. 11, the

projection of Ia on Vt is

Pd / Vt = Ia cos φ = Iq cos δ + Id sin δ

Solving

Iq Xq = Vt sin δ and Id Xd = Eo - Vt cos δ

For Iq and Id , and substituting in (i), gives

Pd = (Eo Vt / Xd ) sin δ + ( Vt2 / 2 ) [ 1/ Xq – 1/ Xd ] sin 2δ per phase ...(iv)

Pd = 3 (Eo Vt / Xd ) sin δ + 3 ( Vt2 / 2 ) [ 1/ Xq – 1/ Xd ] sin 2δ for three phases

The equation can also be established for a salient-pole motor (δ<0), the graph of above

equation is given in Fig. 12. Observe that for Xd = Xq = XS, reduces to the round-rotor

equation.

Fig. 12 Power-angle characteristic of salient-pole machine

30

Parallel Operation of Synchronous Generators

An electric power station often has several synchronous generators operating in

parallel with each other. Some of the advantages of parallel operation are :

1. In the absence of the several machines, for maintenance or some other reason, the

power station can function with the remaining units.

2. Depending on the load, generators may be brought on line, or taken off, and thus

result in the most efficient and economical operation of the station.

3. For future expansion, units may be added on and operate in parallel.

In order that a synchronous generator may be connected in parallel with a system

(or bus), the following conditions must be fulfilled:

1. The frequency of the incoming generator must be the same as the frequency

of the power system to which the generator is to be connected.

2. The magnitude of the voltage of the incoming generator must be the same

as the system terminal voltage.

3. With respect to an external circuit, the voltage of the incoming generator

must be in the same phase as system voltage at the terminals.

4. In a three-phase system, the generator must have the same phase sequence

as that of the bus.

The process of properly connecting a synchronous generator in parallel with a

system is known as synchronizing. Tow generators can be synchronized either by

using a synchroscope or lamps. Figure 1. shows a circuit diagram showing lamps as

well as synchroscope. The potential transformers (PTs) are used to reduce the

voltage for instrumentation. Let the generator G1 be already in operation with its

switch Sg1 closed. Other switches Sg2, S1, and S2 are all open.

After the generator G2 is started and brought up to approximately synchronous

speed, S2 is closed. Subsequently, the lamps La , Lb ,and Lc begin to flicker at a

frequency equal to the difference of the frequencies of G1 and G2. The equality

of the voltages of the two generators is ascertained by the voltmeter V, connected by

the double-pole double-throw switch S. Now, if the voltages and frequencies of the

two generators are the same, but there is a phase difference between the two

31

voltages, the lamps will glow steadily. The speed of G2 is then slowly adjusted

until the lamps remain permanently dark (because they are connected such that

two voltages through them are in opposition). Next, Sg2 is closed and S2 may be

opened.

Fig 1. Synchronizing Two Generators

In the discussion above, it has been assumed that G1 and G2 both have the same

phase rotation. On the other hand, let the phase sequence of G1 be abc

counterclockwise and that of G2 be a'b'c' clockwise. At the synchronous speed of

G1, a and a' may be coincident. This will be indicated by a dark La, but Lb and

Lc will have equal brightness,the phase rotation of G2 must be reversed. When

G2 runs at a speed slightly less than the synchronous speed, with reverse phase

sequence with respect to G1 ,the lamps will be dark and bright in the cyclical

order La,Lb and Lc, the phase rotation of G2 must be reversed with increasing of its

speed to synchronous speed. This process of testing the phase sequence is known as

phasing out.

32

A synchroscope is often used to synchronize two generators which have Previously

been phased out. A synchroscope is an instrument having a rotating pointer,

which indicates whether the incoming machine is slow or fast. One type of

synchroscope is shown schematically in Fig. 2. It consists of a field coil, F,

connected to the main busbars through a large resistance Rf to ensure that the

field current is almost in phase with the busbar voltage, V. The rotor consists of

two windings R and X, in space quadrature, connected in parallel to each other and

across the incoming generator.

Fig 2 A Synchroscope

The windings R and X are so designed that their respective currents are

approximately in phase and 90° behind the terminal voltage, Vi, of the incoming

generator. The rotor will align itself so that the axes of R and F are inclined at an

angle equal to the phase displacement between V and Vi. If there is a difference

between the frequencies of V and Vi, the pointer will rotate at a speed proportional

to this difference. The direction of rotation of the pointer will determine if the

incoming generator is running below or above synchronism. At synchronism, the

pointer will remain stationary at the index. In present-day power stations,

automatic synchronizers are used.

33

Circulating Current and Load Sharing

At the time of synchronizing (that is, when S2 of Fig.1 is closed), if G2 is

running at a speed slightly less than that of G1 the phase relationships of their

terminal voltages with respect to the local circuit are as shown in Fig.3(a). The

resultant voltage Vc acts in the local circuit to set up a circulating current Ic

lagging Vc by a phase angle φc For simplification, if we assume the generators to be

identical, then

tan φc = Ra / Xs

Ic = Vc / 2Zs

Where Ra + J Xs = synchronous impedance , Ra = armature resistance , Xs =

synchronous reactance.

Fig 3 Circulating Currents between Tow Generators

34

Notice from Fig. 3(a) that Ic has a component in phase with V1, and thus acts as a

load on G1 and tends to slow it down. The component of Ic in phase opposition to V2

aids G2 to operate as a motor and thereby G2 picks up speed. On the other hand, if

G2 was running faster than G1 at the instant of synchronization, the phase

relationships of the voltages and the circulating current become as shown in Fig.

3(b). Consequently, G2 will function as a generator and will tend to slow down;

and while acting as a motor, G1 will pick up speed. Thus there is an inherent

synchronizing action which aids the machines to stay in synchronism.

We now recall the power developed by a synchronous machine that Vt is the

terminal voltage, which is the same as the system busbar voltage. The voltage E

is the internal voltage of the generator and is determined by the field excitation.

As we have discussed earlier, a change in the field excitation merely controls the

power factor and the circulation current at which the synchronous machine operates.

The power developed by the machine depends on the power angle δ. For G2 to share

the load, for a given Vt and E the power angle must be increased by increasing the

prime-mover power. The load sharing between two synchronous generators is

illustrated by the following examples.

Example 1: Two identical three-phase wye-connected synchronous generators share equally a

load of 10 MW at 33 kV and 0.8 lagging power factor. The synchronous

reactance of each machine is 6 Ω per phase and the armature resistance is

negligible.If one of the machines has its field excitation adjusted to carry 125 A

of lagging current, what is the current supplied by the second machine ? The

prime mover inputs to both machines are equal.

SOLUTION

The phasor diagram of current division is shown in Fig.4, where in I1= 125 A.

Because the machines are identical and the prime-mover inputs to both machines are

equal, each machine supplies the same true power:

Fig 4

35

I1 cos φ1= I2 cos φ2 = 0.5 I cos φ

I = 10 * 106 / (√3 * 33*10

3 * 0.8) = 218.7 A

I1 cos φ1= I2 cos φ2 = 0.5 * 218.7 * 0.8 = 87.5 A

The reactive current of the first machine is therefore

I1 | sin φ1 | = √ (1252 – 87.5

2) = 89.3 A

And since the total reactive current is

I | sin φ | = 218.7 * 0.6 = 131.2 A

The reactive current of the second machine is

I2 | sin φ2 | = 131.2 – 89.3 = 41.9 A

Hence

I2 = √ (87.52 + 41.9

2) = 97 A

Example 2: Consider the tow machines of example 1. if the power factor of the first machine is

0.9 lagging and the load is shared equally by the tow machines, what are the power

factor and current of the second machine?

SOLUTION

Load:

Power = 10,000 KW, Apparent power =12,500 KVA, Reactive power = 7500 KVar

First machine:

Power = 5000 KW

φ1 = cos-1

0.9 = –25.8º

Reactive power = 5000 tan φ1 = –2422 KVar

Second machine:

Power = 5000 KW

Reactive power = –7500 – (–2422) = – 5078 KVar

tan φ2 = – 5078 / 5000 = – 1.02

cos φ2 = 0.7

I2 = 5000 / (√3 * 33 * 0.7) = 124.7 A

36

Determination of Synchronous Reactance from Open

Circuit and Short Circut Tests

i- Open Circuit Test: The machine is run on no load and the induced

e.m.f. per phase is measured corresponding to various values of field

current and a curve between induced e.m.f. per phase, Eo and field current, If is

drawn which is known as open circuit characteristic (O.C.C.) and has been

illustrated in Fig 5.

Fig 5 Open Circuit Characteristic

ii- Short Circuit Test: The armature winding is short-circuited through a low resistance ammeter. The speed is kept constant during this test and short-circuit current is measured corresponding to various values of field current. The field current or excitation is increased to give short-circuit current about twice the full load current. The short circuit characteristic (S.C.C.) is drawn by plotting a curve between short circuit current Isc as ordinate and field current, If as abscissa, as shown in Fig. 5.

Consider OC the normal field current, then BC gives short circuit current, Isc corresponding to this value of field current on the S.C.C. and AB gives the induced e.m.f. per phase on the O.C.C. for the same excitation. Since on short circuit for field current OC,the whole of the induced e.m.f. AC is utilized to create a short circuit current, Isc given by BC.

Hence synchronous Impedance,

Zs = AC (in volts) / BC (in amperes)

And synchronous reactance,

Xs = √ ( Zs2 – Ra

2)

37

Where Ra is the effective armature resistance per phase which can be measured

directly by volt-meter and ammeter method or by using the wheat-stone bridge.

For normal working conditions the armature resistance measured so is increased by

60% or so. This is being done to allow for skin effect and thus effective armature

resistance Ra is obtained.

Measurement of Xd and Xq

The d-axis synchronous reactance is determined from O.C. and S.C. tests, the q-axis

synchronous reactance can be measured by many methods, one of these methods is

the slip test method.

Slip test method:

From this test, the value of Xd and Xq can be determined. The synchronous machine

is driven by a separate prime-mover (or motor) at a speed slightly different from

synchronous speed. The field windings are left open and positive sequence balanced

voltages of reduced magnitude (around 25% of rated value) and of rated frequency

are impressed across the armature terminals. Under these conditions, the

relative velocity between the field poles and the rotating armature m.m f.

waves is equal to the difference between synchronous speed and the rotor

speed,i.e. the slip speed. A small A.C.voltage across the open field

winding indicates that the field poles and rotating m.m.f. wave, are revolving

in the same direction,and this is what is required in slip test. if field poles

revolve in a direction opposite to the rotating m.m.f. wave, negative sequence

reactance would be measured.

At one instant, when the peak of armature m.m.f. wave is in line with the field

pole or direct axis,the reluctance offered by the small air gap is minimum as shown

in fig. 6 (a). At this instant the impressed terminal voltage per phase divided by

the corresponding armature current per phase, gives d -axis synchronous

reactance Xd. After one-quarter of slip cycle, the peak of armature m.m.f. wave acts on the

interpolar or q-axis of the magnetic circuit, Fig. 6 (b). and the reluctance

offered by long air-gap is maximum. At this instant, the ratio of armature

terminal voltage per phase to the corresponding armature current per

phase, gives q-axis synchronous reactance Xq. Oscillograms of armature

current, terminal voltage and the e.m.f. induced in the open-circuited field

winding are shown in Fig.7. A much larger slip than would be used in practice,

has been shown in Fig.7, merely for the sake of clarity.

38

Fig. 6 pertaining to the physical concepts of (a) Xd , (b) Xq

Fig. 7 Typical oscillograms in slip test

When the armature m.m.f. wave is along the direct axis, the armature flux

passing through open field winding is maximum, therefore, the induced field

e.m.f., i.e. dфa/dt is zero. The d-axis can, therefore, be located on the

oscillogram of Fig.7 where the induced field e.m.f. is zero. When armature

m.m.f. wave is along q-axis, the armature flux linking the field winding is zero,

therefore, the induced field e.m.f. dфa/dt is maximum. Thus the q-axis can also be

located on the oscillogram. If oscillograms can't be taken, then an ammeter

and a voltmeter are used as shown in the connection diagram of Fig.8. The prime-

mover (or d.c. motor) speed is adjusted till ammeter and voltmeter pointers

39

swing slowly between maximum and minimum positions. Under this

condition, maximum and minimum readings of both ammeter and voltmeter are

recorded in order to determined Xd and Xq.

Fig 8 Slip-test connection diagram for obtaining Xd and Xq

Since the applied voltage is constant, the air-gap flux would be constant. When

crest of the rotating m.m.f. wave is in line with the field-pole axis, Fig. 6 (a),

minimum air-gap offers minimum reluctance, consequently the armature

current, required for the establishment of constant air-gap flux, must be

minimum. Constant applied voltage minus the minimum impedance voltage drop

(armature current being minimum) in the leads and 3-phase variac gives

maximum armature-terminal voltage. Thus the d-axis, synchronous reactance is

given by

Xd = (Max. armature terminal voltage per phase) / (Min. armature current per phase)

By a similar thought process

Xq = (Min. armature terminal voltage per phase) / (Max. armature current per phase)

During slip test, it would be observed that swing of the ammeter pointer is

very wide, whereas the voltmeter has only small swing because of the low

impedance voltage drop in the leads and 3-phase variac. Since low armature-

terminal voltages are used, values of reactances obtained are unsaturated values.

When performing this test, the slip should be made as small as possible,

otherwise the currents induced in the amortisseur circuits would cause large

errors in the measurement of Xd and Xq (lower value of reactances for larger

slips). It is however quite difficult to maintain very small slips, as the

reluctance torque due to saliency tends to bring the rotor into synchronism with

the rotating armature m.m.f. wave. It is because of this reason that the slip test

must be conducted at low values of armature terminal voltage so that reluctance

torque due to saliency is low.

40

The advantages of oscillographic method over voltmeter-ammeter methods are

i- elimination of the inertia effects of voltmeter and ammeter

ii- the possibility of large slip-speed, which in turn allows higher armature-

terminal voltages to be applied.

In practice, there may be error in reading the oscillograms. At the same time.

voltmeter ammeter readings are not very reliable because of their inertia effect. In

view of these shortcomings, slip test is conducted only to determine the ratio of

Xd / Xq.

Now, using the value of Xd computed from o.c. and s.c. tests, Xq can be deter-

mined as follow :

Xq = Xd / Xq (from slip test) * Xd (from O.C. and S.C. tests)

41

Voltage Regulation of Alternator

From the previous discussion, it has been known that, the terminal voltage of an alternator changes from no load to full load.

The voltage regulation of an alternator is the difference between the no-load voltage and the full-load voltage exprsed in per cent of full load voltage.

i.e. percentage regulation =( Vno load – V ful l load ) / V ful l load * 100%

= (Eo – Vt) / Vt *100%

Where Eo is no load terminal voltage, Vt is the full load rated terminal voltage

In case of leading power factor, Eo is less than Vt hence the regulation will be

negative.

The regulation of an alternator is usually much higher than that of power transformers. This large regulation results from:

the large amount of leakage reactance present in the alternator.

the armature resistance.

effect of armature reaction (this is the most predominant factor). To make the generator voltage constant, automatic control on the field

current (excitation) is necessary. A change in load will cause readjustment

of excitation in sufficient magnitude so as to provide the constant-potential

characteristic. The device used to control the field current (excitation) is

called a voltage regulator. The regulator changes the alternator from a variable-

voltage machine to a constant- voltage machine.

1. Determination of Voltage Regulation by Synchronous

impedance method or E.M.F. method:

The magnitude of voltage regulation depends upon the load current and power

factor of the load. The vector diagram for any load at any power factor can

be drawn. This has been illustrated in Fig. 1.

Let I be the load current lagging behind the terminal voltage Vt by a phase angle φ.

No load terminal voltage, Eo = OF

Full load terminal voltage, Vt = OA

Armature resistance drop, I Ra= AC, this drop is always in phase with the

load current I.

Synchronous impedance drop, I XS = CF, this drop is always perpendicular to

the load current and also the resistance drop.

42

Fig. 1 Vector Diagram of Alternator

Now

OF2 = OD

2 + DF

2 = ( OB + BD )

2 + ( DC + CF )

2

Or

Eo2 = ( Vt cos φ + I Ra)

2 + ( Vt sin φ + I XS)

2

Since

AC = BD = I Ra And AB = CD = Vt sin φ

Then

Eo = √ [( Vt cos φ + I Ra)

2 + ( Vt sin φ + I XS)

2]

percentage regulation = (Eo – Vt) / Vt *100%

For leading power factor of load, the phase angle φ will be taken as negative and the value Eo will be even less than Vt at large magnitude of leading power factor and the percentage regulation will be negative. For unity power factor taking phase angle φ as zero, the load current will be in phase, with the terminal voltage, Vt .These cases have been illustrated in Fig. 2(a) and 2 (b).

(a) Unity power factor of load (b) Leading power factor of load

Fig 2

The regulation obtained by this method is always higher than the actual

values and is therefore a pessimistic method. However, the results are more

on the safe side and the method is theoretically accurate for non-salient pole

43

machines with distributed field windings when saturation is not considered.

The value of synchronous reactance varies with the saturation. At low

saturation its value is larger because of the effect of armature reaction is

greater than that higher saturation. Under short circuit conditions, saturation

is very low and the value of synchronous impedance (or reactance) measured is

higher than that in actual working conditions.

2. Zero power factor method:

This is also called the general method, Potier reactance (or triangle) method

of obtaining the voltage regulation.

In the e.m.f. method. the phasor diagram involving voltages is used, For the z.p.f

method, the e.m.fs. are handled as voltages and the m.m.fs. as field ampere-turns

or field amperes.

Zero-power-factor characteristic (z.p.f c.), in conjunction with O.C.C., is useful in

obtaining the armature leakage reactance Xl and armature reaction m.m.f. Fa For an

alternator, z.p.f c. is obtained as follows

The synchronous machine is run at rated synchronous speed by the prime-

mover.

A purer inductive load is connected across the armature terminals and field

current is increased till full load armature current is flowing.

The load is varied in steps and the field current at each step is adjusted to

maintain full-load armature current. The plot of armature terminal voltage and

field current recorded at each step, gives the zero -power- factor characteristic

at full-load armature current.

From fig 3(a), it can be seen that the terminal voltage Vt and the air-gap voltage Er are very nearly in phase and are, therefore, related by the simple algebraic equation

Vt = Er – I Xl The resultant m.m.f. Fr and the field m.m.f. Ff are also very nearly in phase and are related by the simple algebraic equation

Ff = Fr + Fa

Assume that O.C.C. gives the exact relation between air-gap voltage Er and the resultant m.m.f. Fr under load. Also, assume armature leakage reactance to be constant.

The O.C.C. and z.p.f c. are shown in Fig.3(b). For field excitation Ff or field current If, equal to OP the open-circuit voltage is PK. With the field excitation and speed remaining unchanged, the armature terminals are connected to a purely inductive load such that full load armature current flows. An examination of Fig. 3(a) and (b) reveals that under zero power. factor load, the net excitation Fr is OF which is less than OP (=Ff)by Fa According to the resultant m.m.f. OF the air-gap voltage Er is FC and if

44

CB=IXl is subtracted from Er= FC, the terminal voltage FB = PA = Vt is obtained. Since z.p.f c. is a plot between the terminal voltage and field current If or Ff, which has not changed from its no-load value of OP, the point A lies on the z.p.f.c. The triangle ABC so obtained is called the Potier triangle, where CB=IXl and BA = Fa. Thus, from the Potier triangle, the armature leakage reactance Xl and armature reaction m.m.f. Fa can be determined.

Fig 3

(a) General phasor diagram of Round-Rotor Alternator at zero power factor

(b) O.C.C., z.p.f c. And potier triangle

If the armature resistance is assumed zero and the armature current is kept constant, then the size of Potier triangle ABC remains constant and can be shifted parallel to itself with its corner C, remaining on the O.C.C. and its corner A, tracing the z.p.f.c. Thus the z.p.f.c. has the same shape as the O.C.C. and is shifted vertically downward by an amount equal to IXl (i.e., ieakage reactance voltage drop) and horizontally to the right by an amount equal to the armature reaction m.m.f. Fa.or the field current equivalent to armature reaction m.m.f.

For determining IXl and Fa experimentally, it is not necessary to plot the entire z.p.f.c. Only two points A and F' shown in Fig 3(b) are sufficimt. The point A(PA=rated voltage) is obtained by actually loading the alternator so that the rated armature current flows in the alternator. The other point F' on the z.p.f.c. corresponds to the zero terminal voltage and can, therefore, be obtained by performing short-circuit test. So here OF' is the field current required to circulate short-circuit current equal to the armature current (generally rated current) at which the point A is determined in the zero(near zero) power-factor test.

45

Now draw a horizontal line AD, parallel and equal to F'O. Through point D, draw a straight line parallel to the air-gap line, intersecting the O.C.C. at C. Draw CB perpendicular to AD. Then ABC is the Potier triangle from which

BC = IXl

AB = Fa

Since the armature current I at which the point A is obtained, is Known, Xl can be calculated.

Then determine the air-gap voltage Er by the relation,

Er = Vt + I ( Ra + j Xl )

According to the magnitude of Er obtain Fr from O.C.C. and draw it leading

Er by 90º, The armature reaction m.m.f. Fa and armature leakage reactance Xl,

can be determined from the Potier triangle, as explained before. Now Fa is

drawn in phase with I as shown in Fig. 3(a) Then

Ff = Fr – Fa

is obtained and corresponding to Ff, excitation voltage Ef (or Eo) is recorded

from O.C.C. and the voltage regulation obtained.

Z.p.f. method requires O.C.C. and z.p.f.c., and gives quite accurate results.

Example 1:

A three phase 6000 V Alternator gave the following open circuit characteristic at

normal speed:

Field current in amper 14 18 23 30 43

Terminal Voltage in volt 4000 5000 6000 7000 8000

With the alternator short circuited and full load current folwing, the field current is

16 A. Neglecting the armature resistance and using synchronous impedance method,

Determine the voltage regulation of the alternator supplying the full load of 200

KVA, at 0.8 power factor lagging.

Solution

The open circuit characteristics have been drawn in fig 4, and corresponding to an

excitation (field) current , If of 16 ampers, the open circuit voltage is 4700 V.

The phase O.C. voltage = 4700 / √3 = 2714 V

Full load current = (200 * 1000) / (√3 * 6000 * 0.8) = 24 A

46

Fig 4

Short curcuit current corresponding to field current of 16 ampers is 24 ampers

Synchronuos impedance ,

Zs = O.C. voltage / S.C. current

= 2714 / 24 = 113 Ω

Now rated voltage of alternator per phase,

V= 6000 / √3 = 3464 volts

And

Cos φ = 0.8

Sin φ = 0.6

Load current, I = 24 ampers

Eo = √ [( Vt cos φ + I Ra)

2 + ( Vt sin φ + I XS)

2]

= √[(3464 * 0.8 + 0)2 + (3464 * 0.6 +24 * 113)

2]

= 5533 V

Therefore,

Percentage regulation at full load = (Eo – Vt) / Vt *100%

= (5533 – 3464) / 3464 *100

= 59.7%

Example 2:

A 220 V, 50 Hz, 6-pole star-connected alternator with, ohmic resistance of 0.06 Ω/ph, gave the following data for open-circuit, short-circuit and full-load zero-power-factor characteristics :

Find the percentage voltage regulation at full-load current of 40 amps at power-factor of 0.8 lag by

(a) e.m f. method (b) z. p. f .

47

Field current, A 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.8 2.2 2.6 3.0

O.C. voltage, V 29 58.0 87.0 116 146 172 194 232 261.5 284 300

S.C. current, A 6.6 13.2 20.0 26.5 32.4 40.0 46.3 59 - - -

z.p.f. terminl voltage, v - - - - - 0.0 29 88 140 177 208

Solution

Rated per phase voltage = 220 / √3 = 127 V

Per phase values for O.C.C. and z.p.f.c. are tabulated below and O.C.C., S.C.C. and

z.p.f.c. are plotted in fig 5.

Field current, A 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.8 2.2 2.6 3.0

O.C. voltage, V 16.7 33.5 50.2 67 84.3 99.3 112 134 151 164 173

z.p.f. terminl voltage,v - - - - - 0.0 16.73 50.8 80.8 102 120

A- E.M.F. Method:

Zs = O.C. voltage / S.C. current

From above tables, Zs ≈ 134 / 59 = 2.27 Ω

Here Xs ≈ Zs = 2.27 Ω, since Ra is quite small.

Eo = √ [( Vt cos φ + I Ra)

2 + ( Vt sin φ + I XS)

2]

= √ [(127 * 0.8 + 40 * 0.06)2 + (127 * 0.6 + 40 * 2.27)

2]

= 196 volts

Percentage regulation = (Eo – Vt) / Vt *100%

= (196 – 127) / 127 *100 = 54%

B- Zero power factor Method:

First of all, the Potier triangle ABC is drawn as described before, Point A corresponds to the rated voltage of 127 V on the Z. p.f.c. The line AD is drawn parallel and equal to F'O = 1.2 A.Then DC is drawn parallel to the air-gap line, meeting the O.C.C. at point C. Perpendicular CB on AD, lives IXl drop equal to 30 volts.

Armature leakage reactance Xl = 30 / 40 = 0.75 Ω

The air-gap voltage Er,

Er = Vt + I ( Ra + j Xl ) = 127(0.8+j0.6) + 40*(0.06+j0.75)= 148.6∟45.6º

Or = √ [( Vt cos φ + I Ra)2 + ( Vt sin φ + I Xl)

2]

=√ [(127 * 0.8 + 40 * 0.06)2 + (127 * 0.6 + 40 * 0.75)

2]

= 148.6 volts

48

Fig 5

Corresponding to Er = 148.6 V, the field current Fr from O.C.C. is 2.134 A, the

armature m.m.f. Fa, from Potier triangle is AB = 0.84 A.

Fr = 2.134 ∟(45.6º+ 90º) = 2.134 ∟135.6º

Fa = 0.84 A

Ff = Fr – Fa = 2.134 ∟135.6º – 0.84

= 2.797 ∟147.7º A

For Ff = 2.797 A, the excitation voltage from O.C.C. is Eo=169.0 volts

Percentage regulation = (Eo – Vt) / Vt *100%

= (169 – 127) / 127 *100 = 33.1%

As already stated, z.p.f. method gives quite accurate results and here the voltage regulation with this method is 33.1%.

The voltage regulation by e.m.f. method is 54% , This value is much higher than the accurate value of 331 % and in view of this, this method may be called pessimistic method.

49

The Automatic Voltage Regulator

The method of operation in the power station is to maintain the terminal voltage of the alternator at the rated value and to adjust the excitation with change in load current accordingly. As the variations in load may be very violent both as regards magnitude and rapidity, it is clear that hand regulation of the excitation is impossible and that automatic means must be adopted. Now the flux per pole of a large turbo-alternator may amount to several webers, and therefore the self-induction of the field winding will be very high. Consider, for example, a 15,000 kVA, 4 pole turbo-alternator having a flux per pole of 1.15 webers, a rotor current of 600 amperes, and a rotor winding of 66 turns per pole. L(per pole)=(Flux per ampere) * (No. of turns)

= (1.15 / 600 )* 66 = 0.127 henry

And Inductance, L = 4 * 0.127 = 0.508 henry for the whole winding

Winding resistance, R=0.2 Ω

The time constant = L / R = 0.508 / 0.2 = 2.54

The time constant L / R of the field circuit gives the value in seconds, and is the time taken for the flux to reach 0.6321 of its final value. With modern large two-pole, turbo-alternators the time constant is considerably greater. What is required in the power station that there shall be an almost instantaneous increase in excitation to the desired value. An increase in load calls for an increase in excitation. This increase is made much greater than is ultimately required, and as a result the flux builds up very rapidly. Before the flux can build up too far, the excitation is reduced again. This is known as the "overshooting-the-mark" principle. There are two types of quick acting regulator which work on this principle : (a) the vibrator type, in which a fixed resistance is rapidly cut in and

out. (b) the rheostat type in which the resistance is variable.

Transistor-Cruntrolled or Transistor Type-Automatic

Voltage Regulators The automatic voltage regulation is affected by matching a quantity proportional to the alternator voltage against a 'reference'. The difference between these two, called 'error' has to be rectified before it can be fed to the excitor, and this is affected by means of transistor amplifiers. This is a 'brushless' method of voltage control for an alternator. Here no slip rings, commutators and brush gear is required. These types of regulators are also called electronic voltage regulators

50

The reference circuit is supplied by voltage feed back from the alternator, the

general scheme of control has been illustrated in block diagram of fig 6.

Fig 6 Block diagram of the control system of a transistor-controlled alternator

51

Synchronous Motor Three Phase

We know that the stator of a three-phase synchronous machine, carrying a

three-phase winding currents produces a rotating magnetic field in the air

gap of the machine. Referring to Fig. 1(a), we will have a rotating magnetic

field in the air gap of the salient pole machine when its stator windings are

fed from a three-phase source. Let the rotor (or field) winding be unexcited.

The rotor will have a tendency to align with the rotating field at all times in

order to present the path of least reluctance. Thus if the field is rotating, the

rotor will tend to rotate with the field. From Fig 1(b) we see that a round

rotor will not tend to follow the rotating magnetic field because the uniform air

gap presents the same reluctance all around the air gap and the rotor does not

have any preferred direction of alignment with the magnetic field. This torque,

which we have in the machine of Fig 1(a), but not in the machine of Fig 1(b), is

called the reluctance torque. It is present by virtue of the variation of the

reluctance around the periphery of the machine.

(a) (b)

Fig. 1 (a) A Salient-rotor machine

(b) A Round-rotor machine

Next, let the field winding [Fig. 1(a) or (b)] be fed by a dc source that produces

the rotor magnetic field of definite polarities, and the rotor will tend to align with

the stator field and will tend to rotate with the rotating magnetic field. We

observe that for an excited rotor, a round rotor, or a salient rotor, both will tend

to rotate with the rotating magnetic field, although the salient rotor will have an

52

additional reluctance torque because of the saliency. In the past lecture we derive

expressions for the electromagnetic torque in a synchronous machine attributable to

field excitation and to saliency.

So far we have indicated the mechanism of torque production in a round-rotor

and in a salient-rotor machine. To recapitulate, we might say that the stator rotating

magnetic field has a tendency to "drag" the rotor along, as if a north pole on the

stator "locks in" with a south pole of the rotor. However, if the rotor is at a

standstill, the stator poles will tend to make the rotor rotate in one direction and then

in the other as they rapidly rotate and sweep across the rotor poles. Therefore, a

synchronous motor is not self-starting. In practice, as mentioned earlier, the

rotor carries damper bars that act like the cage of an induction motor and thereby

provide a starting torque. The mechanism of torque production by the damper bars

is similar to the torque production in an induction motor. Once the rotor starts

running and almost reaches the synchronous speed, it locks into position with the

stator poles. The rotor pulls into step with the rotating magnetic field and runs at

the synchronous speed; the damper bars go out of action. Any departure from the

synchronous speed results in induced currents in the damper bars, which tend to

restore the synchronous speed.

Performance of a Round-rotor Synchronous Motor

Except for some precise calculations, we may neglect the armature resis tance

as compared to the synchronous reactance. Therefore, the steady-state per

phase equivalent circuit of a synchronous machine simplifies to the one shown

in fig 2(a). in which we show the terminal voltage Vt, the internal voltage Eo, and

the armature current Ia, going "into" the machine or "out of" it, depending

on the mode of operation "into" for motor and "out of" for generator .

With the help of this circuit we will study some of the steady-state operating

Fig 2 : (a) an approximate equivalent circuit

(b) power-angle characteristics of a round-rotor synchronous machine

53

characteristics of a synchronous motor. In Fig. 2(b) we show the power-angle

characteristics as given by the power developed equation. Here positive

power and positive δ imply the generator operation, while a negative δ

corresponds to a motor operation. Because δ is the angle between Eo and Vt,

Eo is ahead of Vt in a generator, whereas in a motor, Vt is ahead of Eo. The

voltage-balance equation for a motor is, from Fig. 2(a),

Vt = Eo + j Ia XS

The per phase power developed is

Pd = (Eo Vt / XS ) sin δ

If the motor operates at a constant power,then from above equations required that

Ia XS cos φ = Eo sin δ …(i)

We recall that Eo depends on the field current, If. Consider tow cases:

(1) when If is adjusted so that Eo< Vt, and the machine is underexcited.

(2) when If is increased to a point that Eo > Vt , and the machine becomes

overexcited.

The voltage-current relationships for the two cases are shown in Fig. 3(a).

For Eo > Vt at constant power, δ is greater than the δ for Eo < Vt .Notice that

an underexcited motor operates at a lagging power factor (Ia lagging Vt),

whereas an overexcited motor operates at a leading power factor.

In both cases the terminal voltage and the load on the motor are the same.

Thus we observe that the operating power factor of the motor is

controlled by varying the field excitation, hence altering Eo. This is a very

important property of synchronous motors. The locus of the armature current

at a constant load, as given by (i), for varying field current is also shown in Fig

3(a). From this we can obtain the variations of the armature current Ia with the

field current, If (corresponding to Eo ), and this can be done for different

loads, as shown in figures 3(b)and(c).

54

(c)

Fig 3 (a) phasor diagram for motor operation (Eo', Ia', φ', δ')for underexcited

operation,( Eo'', Ia'', φ'', δ'')for overexcited operation

(b), (c) V-Curves of a synchronous motor

These curves are known as the V- Curves of the synchronous motor. One of the

applications of a synchronous motor is in power factor correction, as demonstrated

by the following examples. In addition to the V curves, we have also shown the

curves for constant power factors. These curves are known as compounding

curves. In the preceding paragraph, we have discussed the effect of change in

the field current on the synchronous machine power factor. However, the load

supplied by a synchronous machine cannot be varied by changing the power factor.

Rather, the load on the machine is varied by instantaneously changing the speed (in

case of a generator, by supplying additional power by the prime mover), and thus

changing the power angle corresponding to the new load. In a synchronous

motor, a load change results in a change in the power angle.

55

EXAMPLE:

A three-phase wye-connected load takes 50 A of current at 0.707 lagging power

factor at 220 V between the lines. A three-phase wye-connected, round-rotor

synchronous motor, having a synchronous reactance of 1.27 Ω per phase. is

connected in parallel with the load. The power developed by the motor is 33

kW at a power angle, δ, of 30°. Neglecting the armature resistance, calculate (a)

the reactive kilovolt-amperes (kVAR) of the motor and (b) the overall power

factor of the motor and the load.

Solution

The circuit and the phasor diagram on a per phase basis are shown in figures

4(a)and(b). From power developed equation, we have

Pd = 1/3 * 33,000 = 220/√3 * Eo/1.27 * sin 30º

Which yield Eo = 220

And from phasor diagram, Ia XS = 127, or Ia = 127/ 1.27 = 100 A, and φa = 30º

The reactive kilovolt-amperes of the motor = √3 * Vt Ia sin φa

=√3* 220/1000 * 100* sin 30 = 19 VAR

Notice that φa is the power-factor angle of the motor, φL is the power angle of the

load, φ is the overall power-factor angle, are shown in fig 4(b). The power angle δ,

is also shown in this phasor diagram, from which

I = IL + Ia

Fig 4 (a) Circuit diagram (b) Phasor diagram

Or algebraically adding the real and reactive components of the currents, we obtain

Ireal = Ia cos φa + IL cos φL

Ireactive = Ia sin φa – IL sin φL

The overall power factor angle, φ, is thue given by

tan φ = [Ia sin φa – IL sin φL ] / [Ia cos φa + IL cos φL] = 0.122

Or φ = 7º and cos φ = 0.992 leading.

56

TORQUE – SPEED curve of a synchronous motor:

In synchronous motor speed remains constant at Ns for all loads as in figure 5.

At any other speed (≠ Ns) ,there is no motor –action , and therefore no torque will

produced.

Fig - 5

Starting of Three Phase Synchronous Motor

As is clear from the construction of the motor, the stator of the synchronous motor

is wound and the winding is connected to a.c. mains while the rotor of the

motor is mostly of salient pole field construction except in special high speed

two pole motors where it is non-salient pole field construction. The rotor is supplied

from a d.c. source.

Consider Fig. 5, the stator for convenience of explanation is shown as having salient

poles N' and S'. When the rotor is excited from d.c. source, there develop N and S

poles on the rotor and this polarity is retained by the rotor throughout but the

polarity of stator poles changes because it is connected to a.c. mains and the

polarity alternates with the frequency of the a.c. supply.

First consider the rotor is stationary and in a position as shown in Fig. 6 (a). At

this instant the similar poles of rotor and stator repel each other and the rotor tends to

rotate in clockwise direction. But half a cycle later (i.e. 1/2f second) the polarityof

the stator poles is reversed [as in Fig 6 (b)] but the polarity of rotor poles remains the

same.

(a) Fig 6 (b)

At this very instant unlike poles of stator and rotor attract each other and, therefore,

the rotor tends to rotate back in the anti-clockwise direction. This shows that the

torque acting on the rotor of the synchronous motor is not unidirectional but

57

pulsating one. It is because of inertia of the rotor, it will not move in any

direction. That is why the synchronous motor is not self-starting one.

Now let the rotor (which is yet unexcited) is speeded up to synchronous or

near synchronous speed by some external arrangement and then it is excited

through the d.c. source. The moment the rotor running at nearly

synchronous speed is excited, it is magnetically locked into position with

the stator poles which runs synchronously and that both in the same

direction. Because of this interlocking between the rotor and stator poles, the

synchronous motor has either to run synchronously or not at all.

Methods of Starting Synchronous Motors

Since the three phase synchronous motor has no starting torque,so artificial

means must be provided for starting it as below:

1. External source. If the d.c. field of a synchronous motor is supplied by a

direct-connected exciter, this exciter can be used as a starting motor. This

method is now rarely used.

2. Induction-motor start. If a squirrel cage winding is constructed in the pole

faces of the synchronous motor (damper bars), it can be used to develop a

starting torque (as well as provide damping) similar to that of the ordinary

induction motor. As the motor reaches about 95 per cent of synchronous speed

(it is operating as an induction motor with 5 percent slip) its field is excited and the

motor pulls into step.

For such a set up, the problem of limiting the starting current without to low a value

of starting torque is met in several ways:

. (a) Auto-transformers are used which are similar in design to those employed on

induction motors. The stator is connected to the reduced voltage supply of the auto-transformer until synchronous or near synchronous speed is reached, and is then connected to the full voltage.

(b) The voltage supplied to the stator can be reduced by using a series reactors

in the supply lines. These reactors give a large voltage drop and low P.F. at

starting.

(c) Various types of rotor windings are used like double squirrel

cage, or using special bar cross-sections (T bar, L bar, or simply deep,

narrow bars) to give high skin effect to the squirrel cage and thus limit

the starting current.

(d) Multiple winding. This involves a special arrangement of stator coils

so that two or more complete windings are paralleled for normal

operation. The paralleled windings have normal values of' reactance.

If one winding is left open during starting process the resistance will

double and the leakage reactance will be greater than the normal value and

will result in a reduced current without the utilization of an auto-

starter. Switching arrangement is shown in Fig. 7.

58

(a) For starting, the short-circuited switch is open and only winding 1 is utilized

(b) The short circuiting switch closes the neutral of the second winding for parallel

operation at normal load

Fig 7 Arrangement of the double-winding synchronous motor

Synchronous Condenser

When a three phase synchronous motor is used for power factor correction with no

mechanical output (no load), it is known as a synchronous condenser. The term

'condenser' applied to device that it draws leading current as does a static

condenser. There is a considerable increase in leading kilovolt- amperes

available when the horsepower load of a synchronous motor is less than its rated

load.

The synchronous condenser is especially designed so that practically all its rated

kVA are available for p.f. correction. Because of the absence of shaft load,

the mechanical design is modified from the ordinary motor standards. Because of

the p.f. adjustment possible, the synchronous condenser is particularly applicable to

transmission line control.

The ability to control the power factor of the synchronous motor by simply

varying the excitation is of very great practical importance. The importance lies in

the fact that the motor can be operated at a leading power factor when desired,

and consequently if the remaining installation has a low power factor, the

overall. power factor for the installation can be brought close to unity.

59

Hunting and Damping of Synchronous Motor

When a synchronous motor is operating under steady-load conditions and

an additional load is suddenly applied, the developed torque is less than

that required by the load and the motor starts to slow down. A slight

reduction in speed increases the phase angle of the generated voltage

and permits more current to flow through the armature Some of the

kinetic energy of the rotating parts is given out to the load during speed

reduction . When the motor is slowing down, it cannot cease deceleration

at exactly the correct torque angle corresponding to the increased load. It

passes beyond this point, develops more than required torque, and

increases in speed. This is followed by a reduction in speed and a

repetition of the entire cycle. Such a periodic change in speed is called

hunting.

The mass of the rotating part and the 'spring effect' of the flux lines are

the necessary elements which give the rotating field structure a natural

oscillating period. If the load varies periodically with this same

frequency or some multiple of it, the tendency is for the amplitude of

these oscillations to increase cumulatively until the motor is thrown

out of step, causing corresponding current and power pulsations. The

mechanical stresses are likely to be severe, and the armature current

greatly increases when the motor leaves synchronism.

A solid pole gives a damping action but has the disadvantage of

excessive pole-face losses unless closed slots are used on the armature.

The most satisfactory arrangement from the standpoint of simplicity

seems to be laminated poles with the squirrel-cage 'amortisseur' or

damping winding. The effectiveness of damper depends upon its

resistance and to a lesser extent upon the length of the airgap. A low

resistance damping winding produces the stronger effect, but if the

synchronous motor is to be started by induction-motor action, using

this squirrel cage, the winding resistance should be fairly high to

produce good starting torque. Because of these opposing tendencies, a

compromise usually must be reached in damping winding design.

60

SHEET 5

SOLVED EXAMPLES IN SYNCHRONOUS MACHINES

Example 1 :

Calculate the percent voltage regulation for a three-phase wye-connected 2500 kVA

6600-V turboalternator operating at full-load and 0.8 power factor lagging. The per

phase synchronous reactance and the armature resistance are 10.4 Ω and 0.071 Ω,

respectively?

Solution:

Clearly, we have XS >> Ra, The phasor diagram for the lagging power factor

neglecting the effect of Ra is shown in Fig. (a). The numerical values are as follows:

Vt = 6600 / √3 = 3810 V

Ia = ( 2500 * 1000 ) / ( √3 * 6600) = 218.7 A

Eo = 3810 + 218.7( 0.8 – j 0.6) j10.4 = 5485∟19.3º

Percentage regulation = ( 5485 – 3810 )/ 3810 *100 = 44%

Example 2 :

Repeat Ex.1 calculations with 0.8 power factor leading as shown in Fig. (b) ?

Solution:

Eo = 3810 + 218.7( 0.8 + j 0.6) j10.4 = 3048∟36.6º

Percentage regulation = ( 3048 – 3810 )/ 3810 *100 = -20%

Fff

H.W. : Repeat Ex.1 calculations with unity power factor , and draw the phasor

diagram for this case ?

61

Example 3 :

A 20-KVA, 220 V, 60 Hz, way-connected three phase salient-pole synchronous

generated supplies rated load at 0.707 lagging power factor.The phase constants of

the machine are Ra = 0.5 Ω and Xd =2 Xq = 4 Ω. Calculate the voltage regulation at

the specified load. ?

Solution:

Vt = 220 / √3 = 127 V

Ia = 20000 / (√3 * 220) = 52.5 A

φ = cos-1

0.707 = 45º

tan δ = (Ia Xq cos φ) / (Vt + Ia Xq sin φ)

= 52.5 * 2 * 0.707 / ( 127 * 52.5 * 2 * 0.707)

= 0.37

δ = 20.6º

Id = 52.5 sin ( 20.6 + 45 ) = 47.5 A

Id Xd = 47.5 * 4 = 190 V

Eo = Vt cos δ + Id Xd = 127 cos 20.6 + 190 = 308 V

Percent regulation = (Eo – Vt ) / Vt *100%

= ( 308 – 127 ) / 127 *100%

= 142 %

Example 4: The stator core of a 4-pole, 3-phase a.c. machine has 36 slots. It carries a short pitch 3-phase winding with coil span equal to 8 slots. Determine the distribution and coil pitch factors?

Solution:

Number of slots per pole = 36 / 4 = 9

Angular displacement between slots = β = 180º / 9 = 20º

Coil span = 20º * 8 =160º

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Therefore, pitch factor = kp = cos (θ/2) , θ = 180º – 160º = 20º

kp = cos 10º = 0.985

Now, number of slots per pole per phase,

m = 9 /3 = 3

Distribution factor,

kd = ( Length of long chord ) / (Sum of lengths of short chords )

= [sin (mβ/2)] / [m sin (β/2)]

= sin 30º / (3 sin 10º) = 0.96

Example 5:

3- phase, 50 Hz generator has 120 turns per phase. The flux per pole is 0.07 weber, assume sinusoidally distributed. Find (a) the e.m f. generated per phase. (b) e.m.f. between the line terminals with star connection. Assume full pitch winding and distribution factor equal to 0.96 ?

Solution:

(a) E.m.f. generated per phase = 4 kf kp k d f T Φp volts

= 4 * 1.11 * 0.96 * 1 * 0.07 * 50 * 120

=1792 volts

(b) E.m.f. between line terminals= √ 3 E.m.f. generated per phase =√ 3 * 1792

= 3100 volts

Example 6:

A three phase, 16-pole, star-connected alternator, has 192 stator slots with eight conductors per slot and the conductors of each phase are connected in series. The coil span is 150 electrical degrees. Determine the phase and line voltages if the machine runs at 375 r.p m. and the flux per pole is 64 mWb distributed sinusoidally over the pole?

Solution:

The flux is sinusoidally distributed over the pole, hence from factor, kf = 1.11

Pitch factor , kp = cos [(180º - 150º) / 2 ] = cos 15º = 0.966

Distribution factor, kd = [sin (mβ/2)] / [m sin (β/2)]

m = Number of slots per pole per phase = 192 / (16 * 3) = 4

β = 180º / (No. of slots/pole) = 180º / (192/16) = 15º

63

kd = [ sin (4 * 15º / 2)] / [4 * sin (15º/2)] = sin 30º / (4 * sin 7.5º) = 0.958

Now total number of conductors per phase

=Total number of slots per phase * number of conductors per slot.

= ( 192 / 3 ) * 8 = 512

Number of turns per phase = T = 512 / 2 = 256

Frequency of generated e.m.f. = f = P N / 120

=[ Number of poles * Synchronous Speed] / 120

=[ 16 * 375] / 120 = 50 cycle per second

Flux per pole = Φp=0.064 wb

Therefore, e.m.f. per phase = 4 kf kp k d f T Φp volts

= 4 * 1.11 * 0.966 * 0.958 * 0.064 * 50 * 256

= 3367 volts

Hence for star connected alternator, Line voltage = √3 * 3367 = 5830 volts

Example 7:

A 2200 volt, star-connected, 50 cycle alternator has 12 poles. The stator has 108 slots each with 5 conductors per slot. Calculate the necessary flux per pole to give 2200 volts on no-load. The winding is concentric and the value of kf can be taken as

1.11 ?

Solution:

Slots per phase = 108 / 3 = 36

Therefore, conductors per phase = 36 * 5 = 180

and number of turns per phase = 180 / 2 = 90

Slots per pole per phase = 36 / 12 =12

Frequency, f =50 cycle per second

kf = 1.11

kp= 1.0 for concentric winding

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kd = [sin (mβ/2)] / [m sin (β/2)]

Where β = 180º / (No. Of slots/pole) = 180º / (108/12) = 20º

m = Number of voltage vectors= Number of slots per pole per phase

= 108 / (12*3) = 3

kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96

Also e.m.f. per phase = Eph = 2200 /√3 volts

The e.m.f. equation for Alternator is

Eph = 4 kf kp k d f T Φp volts

Φp = Flux per pole = Eph / [4 kf kp k d f T ] =( 2200 /√3) / [ 4*1.11*1*0.96*50*90]

= 0.066 wb

Example 8: Find the number of armature conductors in series per phase required for the armature of a 3 phase, 50 Hz, 10 pole alternator with 90 slots. The winding is to be star connected to give a line voltage of 11,000 volts. The flux per pole is 0.16

webers?

Solution:

Number of slots per pole = 90 / 10 = 9

Therefore, β = 180º / 9 = 20º

Slots per pole per phase, m = 9 / 3 = 3

Distribution factor, kd = [sin (mβ/2)] / [m sin (β/2)]

kd = [sin (3*20º/2)] / [ 3 sin (20º/2)] = [sin 30º] / [3 sin 10º] = 0.96

Since there is no mention about the type of winding, the full pitched winding is assumed.

Hence , kp = 1.0

Also , Eph = 11000 / √3 = 6352 volts.

Now , Eph = 4 kf kp k d f T Φp volts

T = [Eph] / [4 kf kp k d f Φp]

= [6352] / [4 * 1.11 * 0.96 * 1 * 0.16 * 50]

= 1863

Number of Armature conductors in serues per phase = 1863 * 2 = 3726

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Example 9 :

A six-pole, 3-phase, 60 cycle alternator has 12 slots per pole and 4 conductors per slot. The winding is 5/6 th pitched. There is 0.025 wb flux entering the armature from each north pole and the flux is sinusoidally distributed along the air gap. The armature coils are all connected in series. The winding is star-

connected. Determine the open circuit e.m.f of the alternatorper phase?

Solution:

Here, number of conductors connected in series per phase

=12 * 6 * 4 / 3 = 96

Therefore, number of turns per phase , T = 96 / 2 = 48

Flux per pole , Φp = 0.025 wb

Frequency, f =50 cycle per second .

kf = 1.11

kp = cos [180º(1 – 5/6) / 2] = cos 15º = 0.966.

Number of slots per pole per phase, m = 12 / 3 = 4

β = 180º / 12 = 15º

Distribution factor ,

kd = [sin (mβ/2)] / [m sin (β/2)]

kd = [sin (4*15º/2)] / [ 4 sin (15º/2)] = [sin 30º] / [4 sin 7.5º] = 0.958

Hence induced e.m.f.,

Eph = 4 kf kp k d f T Φp volts

= 4 * 1.11 * 0.966 * 0.958 * 0.025 * 50 * 48

= 295 volts.