Synchronous Motor

Synchronous Motor-GeneralA synchronous motor is electrically identical with an alternator or ACy y Cgenerator.

A given alternator ( or synchronous machine) can be used as a motor,when driven electrically.

Some characteristic features of a synchronous motor are as follows:

1. It runs either at synchronous speed or not at all i.e. while running itmaintains a constant speed. The only way to change its speed is to varyth l f (b N 120f/P)the supply frequency (because NS=120f/P).

2. It is not inherently self-starting. It has to be run up to synchronous(or near synchronous) speed by some means before it can be(or near synchronous) speed by some means, before it can besynchronized to the supply.

3 It is capable of being operated under a wide range of power factors3. It is capable of being operated under a wide range of power factors,both lagging and leading. Hence, it can be used for power correctionpurposes, in addition to supplying torque to drive loads.

Principle of OperationWhen a 3-phase winding is fed by a 3-phase W e a 3 p ase w d g s ed by a 3 p ase

supply, then a magnetic flux of constant magnitude butrotating at synchronous speed, is produced.

Consider a two-pole stator of Fig. 36.2, in which

Since the two similar poles, N (of rotor) and NS (oft t ) ll S d S ill l h th th t

are shown two stator poles (marked NS and SS) rotating atsynchronous speed say, in clockwise direction.

stator) as well as S and SS will repel each other, the rotortends to rotate in the anti-clockwise direction.

But half a period latter, stator poles, having rotatedaround interchange their position i e NS is at position B andaround, interchange their position i.e. NS is at position B andSS at point A. under these conditions, NS attracts S and SS

attracts N and rotor tends to rotate clockwise.

Hence, we find that due to continuous and rapid rotation of stator poles, therotor is subjected to torque which tends to move it first in one direction and then inthe opposite direction.

Owing to its large inertia, the rotor cannot instantaneously respond to suchquickly reversing torque, with the result that it remains stationary.

Now, consider the condition shown in Fig. 36.3(a) where the statorand rotor poles are attracting each other.

Suppose that the rotor is notstationary, but it is rotatingclockwise, with such a speedthat turns through one pole-pitch by the time the statorpitch by the time the statorpoles interchange theirpositions, as shown in Fig.36 3(b)36.3(b).

Here, again the stator and rotor poles attract each other.

It means that if the rotor poles also shift their positions along with thet t l th th ill ti l i idi ti lstator poles, then they will continuously experience a unidirectional

torque i.e. clockwise torque, as shown in Fig. 36.3.

Method of StartingThere are several methods to start the synchronous motor such as:y

(a) Auxiliary drive (induction motor or dc motor),

(b) Induction start (using damper winding) etc(b) Induction start (using damper winding), etc

General Starting Procedure: The rotor (which is as yetunexcited) is speeded up to synchronous or near synchronous speed byunexcited) is speeded up to synchronous or near synchronous speed bysome arrangement and then excited by the DC source.

The moment this (near) synchronously rotating rotor is excited, it( ) y y g ,is magnetically locked into position with the stator i.e. the rotor poles areengaged with the stator poles and both run synchronously in the samedirection.direction.

It is because of this inter-locking of stator and rotor poles that themotor has either to run synchronously or not at all.

However, it is important to understand that the arrangementbetween the stator and rotor poles is not an absolutely rigid one.

As the load on the motor is increase, the rotor progressively tends tofall back in phase (but not in speed as in DC motors) by some angle(Fi 36 4) b t it till ti t h l(Fig. 36.4) but it still continuous to run synchronously.

The value of this load angle or coupling angle (as it is called)depends on the amount of load to be met by the motor.

In other words, the torque developed by the motor depends on this angle, say, α.

Thi l i f d l d l li lThis angle is referred as load angle or coupling angle.

This is also the angle between rotor and stator fields.

Procedure for Starting a Synchronous Motor Using the Damper WindingUsing the Damper Winding

While starting a modern synchronous motor provided withdamper windings following procedure is adopteddamper windings, following procedure is adopted.

1. First, main field winding is short circuited.2. Reduced voltage with the help of auto-transformers is

applied across stator terminals. The motor starts up.3. When it reaches a steady state speed (as judge by its sound),y p ( j g y ),

a weak DC excitation is applied by removing the short-circuit on the main field winding. If excitation is sufficient,then the machine will be pulled into synchronismthen the machine will be pulled into synchronism.

4. Full supply voltage is applied across stator terminals bycutting out the auto-transformers.

5 Th b d d i d f b5. The motor may be operated at any desired power factor bychanging the DC excitation.

Motor on Load With Constant ExcitationIn a synchronous machine, a back emf Eb (as like DC motor) is set up in the armaturea sy c o ous ac e, a bac e b (as e C o o ) s se up e a a u e(stator) by the rotor flux which opposes the applied voltage V.

This back emf depends on rotor excitation only (and not on speed as in DC motors).

The net voltage in armature (stator) is the vector difference (not arithmetical, as in DCmotors) of V and Eb.

Armature current is obtained by dividing this vector difference of voltages byArmature current is obtained by dividing this vector difference of voltages byarmature impedance (not resistance as in DC machines).

Fig. 36.6 shows the condition when the motor (properly synchronized to the supply) isrunning on no load and has no losses and is having fixed excitation which makesrunning on no-load and has no losses and is having fixed excitation which makesEb=V.

It is seen that vector difference of Eb and V is zeroand so is the armature current.

Motor intake is zero, as there is neither load nor,losses to be met by it.

In other words, the motor just floats.

If motor is on no-load, but it haslosses then the vector for Eb falls backlosses, then the vector for Eb falls back(vectors are rotating anti-clockwise) by acertain angle α (Fig. 36.7), so that a resultant

lt E d h t I i b htvoltage ER and hence current Ia is broughtinto existence, which supplies losses.

If, now, the motor is loaded, then its rotorwill further fall back in phase by a greater

l f l ll d h l d lvalue of angle α-called the load angle orcoupling angle.

Th lt t lt E i i d dThe resultant voltage ER is increased andmotor draws an increased armature current(Fig. 36.8), though at a slightly decreasedpower factor.

Equivalent Circuit of a Synchronous Motor

Fig. 36.9(a) shows the equivalentcircuit model for one armature phaseof a cylindrical rotor synchronous

It is seen from Fig. 36.9(b) that thephase applied voltage V is the vector sumof reversed back emf i.e. –Eb and theof a cylindrical rotor synchronous

motor.b

impedance drop IaZs.

In other words, V=(-Eb+IaZs).

The angle α between the phasor for V and Eb is called the load angle or powerangle of the synchronous motor.

Power Flow within a Synchronous MotorLet Ra = armature resistance/phase;e a a a u e es s a ce/p ase;

XS = synchronous reactance/phaseThen, ZS=Ra+jXS;

Hence, the power available at the shaftwould be less than the developedpower by this amount.

Sb

SRa Z

EVZEI -

==

b ZIEVObviously

Out of the input power/phase VIacosφ,and amount Ia

2Ra is wasted inarmature, the rest (VIacosφ- Ia

2Ra)Sab ZIEV −=Obviously,

The angle θ (known as internal angle) bywhich Ia lags behind ER is given by

θ / f i li ibl h θ 90

a a aappears as mechanical power in rotor;out of it, iron, friction and excitationlosses are meet and the rest is available

tanθ=XS/Ra. If Ra is negligible, then θ=90o.

Motor input = VIacosφ-per phase, here V isapplied voltage/phase.

at the shaft.

If power input/phase of the motor is Pthen: P=Pm+ Ia

2Ra.pp g p

Total input for a star-connected 3-phasemachine is, P=√3VLILcosφ

m a a

Or mechanical power in rotor Pm=P-Ia

2Ra -per phase

√The mechanical power developed, somewould go to meet iron and fraction andexcitation losses.

For three phases Pm=√3VLILcosφ-3Ia

2Ra.

The per phase power development in a synchronous machine is as under:

Power Developed by a Synchronous MotorThe armature resistance of a synchronous motor is negligible as compared to itsThe armature resistance of a synchronous motor is negligible as compared to itssynchronous reactance.

Hence the equivalent circuit for the motor becomes as shown in Fig. 36.10 (a).

From the phasor diagram of Fig. 36.10(b), it is seen that φ== cossin SXaIbEAB α

αsincosOr, VbESXaVI =φ

Now, VIacosφ = motor power input/phase

phasepersin −=∴ αSXVbE

inP phasethreeforsin3−=∴ α

SXVbE

inP

Since Stator Cu losses have been neglected P also 3 VESince Stator Cu losses have been neglected, Pin alsorepresents the gross mechanical power (Pm)developed by the motor.

αsin3

SXVbE

inPmP ==∴

The gross torque developed by the motor is T =9 55P /N N m N in rpmThe gross torque developed by the motor is Tg=9.55Pm/NS N-m -NS in rpm.

mNsin65.28sin355.955.9 −=×

==∴SXSN

VbE

SXSNVbE

SNmP

gT αα

Example 38.1 A 75 kW, 3-phase, Y-connected, 50 Hz, 440 V cylindrical rotorsynchronous motor operates at rated condition with 0.8 power factor leading. Themotor efficiency excluding field and stator losses, is 95% and XS=2.5 ohm. Calculate:o o e c e cy e c ud g e d a d s a o osses, s 95% a d S .5 o . Ca cu a e:(i) mechanical power developed, (ii) armature current, (iii) back emf, (iv) power angle,and (v) maximum or pull-out torque of the motor.

S l i N 120 50/4 1500 25Solution: NS= 120×50/4= 1500 rpm = 25 rps.W950,7895.0/10375/)( =×=== ηoutPmPinPi

(ii) Si i t i k k th t(ii) Since power input is known, we know that

A1298.04403

950,78W;950,788.04403cos3 =××

==×××=φ∴ aIaIaILV

(iii) A li d l / h 2 44403 2 4∠0 h i i 36 11(iii) Applied voltage/phase= 2544403 =× Let V= 254∠0o as shown in Fig. 36.11.°=−=φ+= 9.36)8.0(1cosandNow, SjIXbEV

°−∠=°∠×°∠−°∠=−= 30516905.29.361290254or, SjIXVbE Sjb°−=∴ 30,anglePower(iv) α

(v) Maximum or Full-out torque occurs when α=90o.

W8.276,15790sin5.2

5162543sin3(max) =°

××== α

SXVbE

mP∴Pull-out torque=9.55×157,276.8/1500=1001.328 N-m

Synchronous Motor with Different ExcitationsA synchronous motor is said to have normal excitation when its Eb=V.

If field excitation is such that Eb<V, the motor is said to be under-excited.

In both these conditions, it hasa lagging power factor asshown in Fig. 36.12.

If DC field excitation is such that Eb>V, then motor is said to be over-excited and

g

If DC field excitation is such that Eb V, then motor is said to be over excited anddraws a leading current as shown in Fig. 36.13(a)

Th ill b l fThere will be some value ofexcitation for which armaturecurrent will be in phase with V, sothat power factor will become unitythat power factor will become unity,as shown in Fig. 36.13(b).

The value of α and back emf Eb can be found with the help of vectordiagrams for various power factors.(i) L i P F A f Fi(i) Lagging Power Factor: As seen from Fig. 36.14(a) φ−θ+φ−θ−=+= 2)]sin([2)]cos([222

REREVBCABAC

⎞⎜⎜⎛⎞

⎜⎛ φ−θ−−

φ−θ+φ−θ−==

)sin(1tan1tan

2)]sin([2)]cos([

SZaIBC

SZaISZaIVACbE

αaRsX1tan−=θ

bR += EVE

⎠⎜⎜⎝⎠

⎜⎝ φ−θ−

==)cos(

tantanSZaIVAB

α

)cos(222 φθ +−+= RVEREVbE

)sin]cos[)]o180sin()o180[cos()o0sino0(cos

αααα

bjEbEVR

jbEjVRbR

+−=−+−++=

EE

Rb

αsin:isandbetweenAngle

bEVRE

αα

φθ cossin

1tan-bEV

bE−

−=

(ii) Leading Power Factor: As seen from Fig. 36.14(b)22

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

φ−θ−°−φ−θ−°−=

φ+θ−°+φ+θ−°+=

)}(180cos{)}(180sin{1tan

2)}](180sin{[2)}](180cos{[

SZaIVSZaI

SZaISZaIVbE

α

Where, α=load angle

⎠⎝

⎞⎜⎛ φ+θ−=

φ+θ+φ+θ−=

)sin(1tan

2)]sin([2)]cos([

SZaISZaISZaIVbE

α⎠

⎜⎜

⎝ φ+θ−= )cos(tanSZaIV

Sα

)cos(222 φθ −−+= RVEREVbE

)]o180sin()o180[cos()o0sino0(cos αα jbEjVRbR

−+−++=

+=

EEVE

)sin]cos[)]()[ ()(

αα bjEbEVR

jbjR+−=E

:anglefactor Power

aRsX1tan

g−=θ

(iii) Unity Power Factor: As seen from Fig. 36.14(c)

lfP

aRsX1tan

:anglefactor Power−=θ

Here, OB=IaRa and BC=IaXS;

aR

⎞⎜⎜⎛−=++= SXaI

SXaIaRaIVbE 1tan;2][2][ α⎠

⎜⎜⎝ −

++aRaIVSXaIaRaIVbE tan;][][ α

Example 20.1 A 25-hp 220V 60Hz four pole Y-connectedsynchronous motor is rotating with a light load. The angle betweeny g g gthe rotor and stator fields is 3o. The excitation is adjusted for agenerated armature voltage per phase of 110V. (a) what is theresultant armature voltage per phase? (b) what is the angle betweenresultant armature voltage per phase? (b) what is the angle betweenresultant voltage (ER) and terminal voltage (V).

Solution: Terminal voltage per phase, V=220/√3=127Vg p p , √

α= 3o. Cosα =0.9986 and sin α=0.0523

We know that )i][ jEEVEWe know that, )sin]cos[ αα bjEbEVR +−=E

2)]sin[2]cos[)(Thus, αα bEbEVREa +−=

VRE 95.172]0523.0110[2]9986.0110127[ =×+×−=

i o7.189986.0110127

0523.01101tan

cossin

1tan-)( =×−

×−=

−−=

αα

φθbEV

bEb

Effect of Increased Load with Constant ExcitationThe effect of increased load on a synchronous motor will be studied under

diti f l d d it ti (i i th ff t f tconditions of normal, under and over-excitation (ignoring the effects of armaturereaction).

With normal-excitation, Eb=V, with under-excitation, Eb<V and with over-b bexcitation, Eb>V. The value of excitation would be kept constant.

Ra is considered negligible as compared to XS so that phase angle between ER and Iai.e. θ=90o.i.e. θ 90 .

Normal Excitation, (Eb=V )Fig. 36.15(a) shows the condition when motoris running with light load so that (a) torqueangle α1 is small, (b) so ER1 is small, (c) henceIa1 is small and (d) φ1 is small so that cosφ1 islarge.

Now, suppose that load on the motor is increased as shown in Fig. 36.15(b)., pp g ( )

For meeting this extra load, motor must develop more torque by drawing morearmature current.

What actually happens is as under:

1. Rotor falls back in phase i.e. load angle increases to α2 as shown inFig. 36.15(b)

2 Th lt t lt i t i i d id bl t2. The resultant voltage in armature is increased considerably to newvalue ER2.

3 As a result I increases to I thereby increasing the torque3. As a result, Ia1 increases to Ia2, thereby increasing the torquedeveloped by the motor.

φ1 increase to φ2, so that power factor decreases from cosφ1 to the newφ1 increase to φ2, so that power factor decreases from cosφ1 to the new value cosφ2.

Since increase in Ia is much greater than the slight decrease inpower factor, the torque developed by the motor is increased (on thewhole) to a new value sufficient to meet the extra load put on thewhole) to a new value sufficient to meet the extra load put on themotor.

It will be seen that essentially it is by increasing its I that theIt will be seen that essentially it is by increasing its Ia that themotor is able to carry the extra load put on it.

The effect of increased load on a synchronous motor at normalexcitation is shown in Fig. 36.16.

It is seen thatthere is comparativelym ch greater i inmuch greater increase inIa than in φ.

Under-excitation (Eb<V )As shown in Fig 36 17 with aAs shown in Fig. 36.17, with asmall load and hence, small torqueangle α1, Ia1 lags behind V by alarge phase angle φ1 which meanspoor power factor.

A much larger armature current must flow for developing the samepower because of poor power factor.

A l d i E i t E tl I i tAs load increases, ER1 increases to ER2, consequently Ia1 increases toIa2 and power factor angle decreases from φ1 to φ2.

Due to increase both in I and power factor power generated by theDue to increase both in Ia and power factor, power generated by thearmature increases to meet the increased load.

As seen in this case change in power factor is more than the changeAs seen, in this case, change in power factor is more than the changein Ia.

Over-excitation (Eb>V )When running on light load α is small but I is comparatively largerWhen running on light load, α1 is small, but Ia1 is comparatively largerand leads V by a larger angle φ1.

Like the under excited motor as motor load is applied the power factorLike the under excited motor, as motor load is applied, the power factorimproves and approaches unity.

The armature current also increases thereby producing the necessaryy p g yarmature power to meet the increased load (Fig. 36.18).

I thi f t l φIn this case, power factor angle φdecreases (or power factorincreases) at a faster rate than thearmature current therebyproducing the necessary increasedpower to meet the increased loadpower to meet the increased loadapplied to the motor.

The main points regarding the above three cases can be summarizedas under:

1. As load on the motor increases, Ia increases regardless ofexcitation.

2. For under- and over-excited motors, power factor tends toapproach unity with increase in load.

3 B h i h d d i i h i f i3. Both with under- and over-excitation, change in power factor isgreater than Ia with increase in load.

4 With normal excitation when load is increased change in I is4. With normal excitation, when load is increased change in Ia is greater than in power factor which tends to become increasingly lagging.

Different Torques of a Synchronous MotorVarious torques associated with a synchronous motor are asVarious torques associated with a synchronous motor are as follows:

1. Starting torque,

2. Running torque,

3 P ll i t d3. Pull-in torque, and

4. Pull-out torque.

(a)Starting Torque: It is the torque developed by the motorh f ll l i li d i ( )when full voltage is applied to its stator (armature)

winding.

It is also sometimes called breakaway torque.

(b) Running Torque: As its name indicates, it is the torque developedby the motor under running conditions.

It is determined by the horse-power and speed of the drivenmachine.

The peak horse power determines the maximum torque that wouldbe required by the driven machine.

The motor must have a breakdown or a maximum running torquegreater than this value in order to avoid stalling.

(c) Pull-in Torque: A synchronous motor is started as induction motortill it runs 2 to 5% below the synchronous speed.

Af d i i i i h d d h ll iAfterwards, excitation is switched on and the rotor pulls into stepwith the synchronously rotating stator field.

The amount of torque at which the motor will pull into step isThe amount of torque at which the motor will pull into step iscalled the pull-in torque.

(d) Pull-out Torque: The maximum torque which the motor candevelop without pulling out of step or synchronism is called the pull-

t tout torque.

Normally, when load on the motor is increased, its rotorprogressively tends to fall back in phase by some angle (called loadprogressively tends to fall back in phase by some angle (called loadangle) behind the synchronously-revolving stator magnetic fieldthough it keeps running synchronously.

Motor develops maximum torque when its rotor is retarded by anangle of 90o.

Any further increase in load will cause the motor to pull out of step(or synchronism) and stop.

Effect of Excitation on Armature Current and Power FactorCurrent and Power Factor

Consider a synchronous motor in which the mechanical load isconstantconstant.

The value of excitation for which back emf Eb is equal (inmagnitude) to applied voltage V is known as 100% excitation whichmagnitude) to applied voltage V is known as 100% excitation whichis shown in Fig. 36.47(a).

The armature current I lags behind V by a small angle φThe armature current I lags behind V by a small angle φ.

Its angle θ with ER is fixed by stator constants i.e. tanθ=XS/R.

In Fig. 36.47(b) excitation is less than 100%i.e. Eb<V.b

Here, ER is advanced clockwise and so isarmature current (because it lags behind ERarmature current (because it lags behind ERby fixed angle θ).

The magnitude of I is increased but itsgpower factor is decreased (φ has increased).

Because input as well as V are constant hence the powerBecause input as well as V are constant, hence the powercomponent of I i.e. Icosφ=OA will remain constant.

When field current is reduced the motor pull-out torque is alsoWhen field current is reduced, the motor pull out torque is alsoreduced in proportion.

Fig. 36.47(c) represent the condition for over-excited motor i.e. Eb>V.Here the resultant voltage vector E is pulled anti-clockwise and so isHere, the resultant voltage vector ER is pulled anti-clockwise and so isI.Now motor is drawing a leading current.

It may also happen for somevalue of excitation, that I may be, yin with V i.e. power factor isunity as shown in Fig. 36.47(d).

At that time, the current drawn by the motor would be minimum.

Two important points stand out clearly from the above discussion:

1. The magnitude of armature current varies with excitation.

The current has large value both for low and high values ofexcitation.b i h

ΙΙΙIn between, it hasminimum valuecorresponding to a

Ι ΙΙΙΙΙ

p gcertain excitation.

The variations of I withexcitation are shown inFig. 36.48(a) which areknown as ‘V’ curves

Ι. No load

ΙΙ. Half-load

ΙΙΙ F ll L dknown as V curvesbecause of their shape.

ΙΙΙ. Full Load

2. For the same input, armature current varies over a wide range and socauses the power factor also to vary accordingly.

When over-excited, motor runs with leading power factor and withlagging power factor when under-excited.

In between, the power factor inunity.

Th i i f fThe variations of power factorwith excitation are shown inFig. 36.48(b).g ( )

The curve for power factorlooks like inverted ‘V’ curve.

It would be noted that minimumarmature current corresponds

i fto unity power factor.

It is seen that an over-excited motor can be run with leading powerfactor.

This property of the motor renders it extremely useful for phaseadvancing (and so power factor correcting) purposes in the case ofi d t i l l d d i b i d ti t (Fi 36 49) d li hti

B th t f d i d ti t d l i t f th

industrial loads driven by induction motors (Fig. 36.49) and lightingand heating loads supplied through transformer.

Both transformers and induction motor draw lagging currents from theline.

The power drawn by them has a large reactive component and thepower factor has a very low value.

This reactive component, though essential for operating the electricalmachinery, entails appreciable loss in many ways.

The lagging reactive power of induction motor and transformersrequired is supplied by synchronous motors.

The synchronous motors supply only the active component of the loadcurrent.

When used in this ways, a synchronous motor is called a synchronouscapacitor (synchronous condenser), because it draws, like a capacitor,leading current from the line.g

Advantages of SM over IM

Synchronous motors have the followingadvantages over induction motor:

1. SM can be used for power factor correction inaddition to supplying torque to drive loadsaddition to supplying torque to drive loads

2. SMs are more efficient (at unity power factor)than IM of corresponding horsepower andvoltage rating.

3. The field pole rotor of SM can permit the useof wider airgap than the squirrel cage IM.of wider airgap than the squirrel cage IM.

Advantages of SM over IM continue………..

4 SM can operate lagging leading and unity4. SM can operate lagging, leading and unity power factor but IM can operate only lagging power factorpower factor.

5. SM may less costly for the same horsepower, speed and voltage ratingsspeed and voltage ratings.

6. They give constant speed from no-load to full-loadload.

Disadvantages of SM over IM

1. They require dc excitation which must bey qsupplied from external source.

2. They have tendency to hunt.y y3. They cannot be used for variable speed jobs

as speed adjustment cannot be done.p j4. They cannot be started under load. Their

staring torque is zero.g q5. They may fall out of synchronism and stop

when overloaded.

Compare a Synchronous Motor with an Ind ction MotorInduction Motor

1 For a given frequency the synchronous motor runs at constant1. For a given frequency, the synchronous motor runs at constantaverage speed whatever the load, while the speed of an inductionmotor falls somewhat with increase in load.

2. The synchronous motor can be operated over a wide range ofpower factors, both lagging and leading, but induction motorl ith l i f t hi h balways runs with a lagging power factor which may become very

low at light loads.

3 A synchronous motors is inherently not self starting but induction3. A synchronous motors is inherently not self-starting but inductionmotor is self-starting.

4 The changes in applied voltage do not affect synchronous motor4. The changes in applied voltage do not affect synchronous motortorque as much as they affect the induction motor torque.

5. The breakdown torque of a synchronous motor variesq yapproximately as the first power of applied voltage whereas that ofan induction motor depends on the square of this voltage.

6. A DC excitation is required be synchronous motor but not byinduction motor.

h ll l d li d h7. Synchronous motor are usually more costly and complicated thaninduction motors, but they are particularly attractive for low speeddrives (below 300 rpm) because their power factor can always be( p ) p yadjusted to 1.0 and their efficiency is high. However, inductionmotors are excellent for speeds over 600 rpm.

8. Synchronous motor can be run at ultra-low speeds by using highpower electronics converters which generated very lowfrequencies. Such motors of 10 MW range are used fro drivingfrequencies. Such motors of 10 MW range are used fro drivingcrusher, rotary kilns and variable speed ball mills etc.

Synchronous Motor Applications

Synchronous motors find extensive application forSynchronous motors find extensive application forthe following classes of service:

1. Power factor correction;

2 Constant speed constant load drives; and2. Constant speed, constant load drives; and

3. Voltage regulation.

The synchronous motors have the following fields ofapplications:applications:

Power houses and sub-station: Used in power houses andsub-station in parallel to the bus-bar to improve thepower factor.

Factories: Used in factories having large of inductiont th t ti tmotors or other power apparatus, operating at

lagging power factor, to improve the power factor.Mills-industries etc : Used in textile mills rubber millsMills-industries etc.: Used in textile mills, rubber mills,

and other big industries, cement factories for powerapplications.pp

Constant speed equipments: Used to drive continuouslyoperated and constant speed equipment such as:

ifFans, blowers, centrifugal pumps, motor-generatorsets, and air compressors etc.

Power Factor CorrectionIt has been seen that a synchronous motor can be operate in bothlagging and leading power factor by varying the field flux ofexcitation circuitexcitation circuit.

An over-excited (back emf, Eb > supply voltage, V) motor can be runwith leading power factorwith leading power factor.

When a synchronous motor operates at leading power factor, thebehavior of its as like a capacitor or condenser. That’s why abe av o o s as e a capac o o co de se . a s w y asynchronous motor operated at leading power factor is referred to asynchronous capacitor (synchronous condenser).

Thus, over-excited synchronous motors leading power factor arewidely used for improving power factor (by supplying leading reactivepower) of those power systems which employ a large number ofpower) of those power systems which employ a large number ofinduction motor and other devices having lagging power factor.

Constant Speed, Constant Load Drives

Because of their high efficiency and high speed,synchronous motors (above 600 rpm) are well suited forloads where constant speed is required such as centrifugal

b l d i i i bl lipumps, belt-driven reciprocating compressors, blowers, lineshafts, rubber and paper mills etc.

Low-speed synchronous motors (below 300 rpm) are used for drives such as centrifugal and screw type pumps, b ll d t b ill hi d t lball and tube mills, vacuum pumps, chippers and metal rolling mills.

Voltage regulationThe voltage at the end of along transmission line variesThe voltage at the end of along transmission line variesgreatly especially when large inductive load are present.When an inductive load is disconnected suddenly, voltagey, gtends to rises considerably above its normal value because ofthe line capacitance. By installing synchronous motor with afield regulator (for varying its excitation), this voltage risecan be controlled.

When line voltage decreases due to inductive load, motorexcitation is increased thereby raising its power factor which

f h li d f h h h d licompensates for the line drop. If, on the other hand, linevoltage rise due to line capacitive effect, motor excitation isdecreased thereby making its power factor lagging whichdecreased, thereby making its power factor lagging whichhelps to maintain the line voltage at its normal value.

Example 38.31 A synchronous motor absorbing 60 kW is connected inparallel with a factory load of 240 kW having a lagging power factor of0 8 If the combined load has a power factor of 0 9 what is the value of0.8. If the combined load has a power factor of 0.9, what is the value ofthe leading kVAR supplied by the motor and what power factor is itworking?Solution: Load connections and phase relationships are shown in Fig.36.53.

T t l l d 240+60 300 kW bi d f t 0 9 (l )Total load= 240+60 = 300 kW; combined power factor =0.9 (lag);

φ=25.8o; tanφ=0.4834; combined kVAR=300×0.4834=145 (lag)

Factory Load: cosφL=0.8; φL=36.9o; tanφL=0.75;

Load kVAR=240×0 75=180 (lag)Load kVAR 240×0.75 180 (lag)

[or load kVA=240/0.8=300, kVVAR=300×sinφL=300×0.6=180]

l di kVAR li d b h 180 145 35∴leading kVAR supplied by synchronous motor =180-145=35

For Synchronous Motor: kW= 60; leading kVAR=35; tanφm=35/60;φ =30 3o; cos30 3o=0 863;φm=30.3o; cos30.3o=0.863;

∴ Motor power factor=0.863 (leading).22Incidentally, motor kVA= 5.69235260 =+