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ENGNG 2024 Electrical Engineering

E Levi, 2000 1

SYNCHRONOUS MACHINES

1. INTRODUCTION

Synchronous machines come in a variety of different constructions and designs. Thedifferences occur in the physical outlay of the rotor and in the way in which excitation flux isprovided (if it is provided at all) in the machine. Regardless of the type however, all thesynchronous machines have the same construction of the stator. Stator is of cylindrical cross-section, manufactured from laminated sheets of steel, and it carries a three-phase winding thatis supplied with (in the motor case) or that produces (in the generator case) a system of three-phase voltages. The three windings that constitute the stator three-phase winding are displacedin space by 120 degrees around the circumference of the machine. Three-phase voltages havethe phase displacement of 120 degrees.

Synchronous machines are the main work-horse of the electricity generation industry.They are used as generators in all the hydro, nuclear, coal-fired, gas-fired and oil-fired powerplants. This means that a synchronous generator is a standard machine used for conversion of mechanical energy into electric energy in all the power plants that rely on conventional energysources. Rated powers of synchronous generators are typically from a few megawatts up to afew tens of megawatts, or even a several hundreds of megawatts. Synchronous machines areused as motors as well. In this case rated power of a synchronous motor is either relativelyvery low (up to few kilowatts) or is in the high power region, from around 150 kW to 15 MW.In between, induction motors are used as a rule, due to their numerous advantages in thispower region.

The two types of synchronous machines that are relevant for the discussion that followsare:

1. Machines with uniform air-gap and excitation winding on rotor.

2. Machines with non-uniform air-gap and excitation winding on rotor.

Rotor cross-section in these two types differs and leads to different mechanisms of torque production. Both machine types are used for both generating and motoring application.In the machine with uniform air-gap torque is produced solely due to interaction between therotor and stator windings (fundamental torque component). In the machine with non-uniformair-gap, there are two torque components: fundamental torque, produced by the interactionbetween the stator and rotor windings, and reluctance torque component that is theconsequence of the non-cylindrical rotor cross-section. Since the rotor is not cylindrical, statorwindings see an air-gap of variable length as rotor rotates. Consequently, magnetic reluctanceis variable and all the inductances of the stator windings are functions of the rotor position.Reluctance torque component is therefore produced, in addition to the fundamental torquecomponent.

Synchronous machine with uniform air-gap and an excitation winding on rotor:

Rotor is of cylindrical construction, with an excitation winding placed on rotor. Theexcitation winding is supplied with a controllable DC current. Since the rotor rotates, this DC

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E Levi, 2000 2

current has to be supplied somehow from the stationary outside world. In the past, slip ringsand brushes were used to connect a stationary DC source to the rotating rotor winding.Nowadays, so-called brushless excitation systems (or just brushless exciters) are used (theseare beyond the scope of interest here). Synchronous machines with uniform air-gap andexcitation winding on rotor are often termed turbo-machines. They are used both for motoring

application and for generation in coal-, oil-, and gas-fired plants and nuclear plants. Typicalnumber of magnetic pole pairs is one or two, so that the so-called synchronous speed of rotation is 3000 rpm or 1500 rpm for 50 Hz stator frequency. Synchronous speed is defined as

n f P f Ps s= =60 2 / / (1)

and this is the so-called mechanical synchronous speed of rotation (that is, actual speed of rotation). Symbol P stands for the number of magnetic pole pairs (each pole pair consists of two magnetic poles, one north and one south).

Rotor of a turbo-machine carries, apart from the excitation winding, a short-circuitedmulti-phase winding similar to the squirrel-cage winding of an induction motor. The existence

of this winding is crucial for motoring applications since, without it, the machine would not becapable of starting when connected to a three-phase supply on stator side. It is importanthowever to emphasise that this winding has no impact on steady-state operation, since nocurrents exist in the short-circuited rotor winding (often called damper winding) in steady-state. The damper winding is important in generation as well, since it has an important roleduring transients. Once again, however, it has no impact on steady-state operation and istherefore omitted from the schematic representation of a synchronous turbo-machine in Fig. 1,where only excitation winding is shown on rotor. Stator three phase winding is shown in Fig. 1with three concentric turns.

It is important to stress once more that an electromagnetic torque in a synchronous

machine with uniform air-gap is produced purely due to the interaction between the statorcurrents and the rotor flux (that stems from the DC current in the rotor winding). The torqueconsist therefore entirely of the fundamental torque component, very much the same as thecase is with other uniform air-gap machines (DC machines, induction machines). According tothe condition of average torque existence, zero frequency in rotor means that the statorfrequency must equal the frequency of rotation. Rotor flux, produced by rotor DC current, isstationary with respect to rotor. However, as rotor rotates at the frequency of stator supply,rotor flux seen from the stator rotates at the same, synchronous, speed as does the statorrotating field created by the system of stator three-phase currents. Hence all the fields in asynchronous machine rotate at the same, synchronous speed. Moreover, rotor rotates at thesynchronous speed as well. Note that for the constant frequency of stator voltages there is asingle speed at which the machine can rotate. This means that, regardless of the loading, asynchronous machine rotates at a constant speed. This is in huge contrast to some other typesof electric machines (DC machines, induction machines) where the speed of rotation isdependent on the loading of the machine.

Synchronous machine with non-uniform air-gap and excitation winding on rotor:

As already noted, if the air-gap is not uniform, magnetic reluctance (resistance) seen bywindings will vary. In synchronous machines with non-uniform air-gap stator remains to becylindrical, but the rotor is not. Consequently, magnetic reluctance seen by stator windings willvary as the rotor rotates. This gives rise to dependence of stator inductances on instantaneousrotor position and leads to creation of the second torque component, not present in uniform

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E Levi, 2000 3

air-gap machines (reluctance torque component). Since the machine contains an excitationwinding, then a fundamental torque component exists as well, so that the total torque is a sumof the fundamental torque component and the reluctance torque component.

Rotor b -c StatorAir-gap

-a a

c -b Rotorwinding

Fig. 1 - Synchronous machine with uniform air-gap and excitation winding on rotor.

The structure of the machine is shown in Fig. 2. The rotor is of so-called salient-polestructure and the machine is usually used for very low speed of rotation, so that the number of poles is very high (say, 84 poles - that is, 42 pole pairs, is a rather common structure). Themachine of Fig. 2 is shown as being of 2-pole structure for simplicity. This type of synchronousmachines is often called hydro-machine or salient-pole synchronous machine, since thesemachines are used in hydro-powered electric generation plants. The machine is used in

motoring application as well, for those cases where a low speed of rotation is required.Rotor of a hydro-machine carries, apart from the excitation winding, a short-circuited

multi-phase winding similar to the squirrel-cage winding of an induction motor (this winding isnot shown in Fig. 2). This is very much the same as for a turbo-machine. This winding oncemore has no impact on steady-state operation, since no currents exist in the short-circuitedrotor winding in steady-state (there is not any induced emf, since the rotor rotates at the samespeed as does the stator rotating field).

Rotorb -c Stator

Air-gap

-a a

Rotorc -b winding

Fig. 2 - Cross-section of a salient pole synchronous machine with excitation winding.

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E Levi, 2000 4

As already noted, synchronous machines are used as motors either for low powerapplications or for very high power applications. In the low power range the advantage of synchronous motors is their fixed synchronous speed of rotation that is always governed by theapplied frequency in stator winding. In other words, speed of rotation depends only on statorfrequency and it is load independent, in contrast to DC motors and induction motors. As far as

the high power region is concerned, the advantage of synchronous motors over the inductionmotors is that they are capable of producing reactive power (while induction motors alwayshave to consume reactive power). The higher capital cost of a synchronous motor, whencompared to an induction motor of the same size, is thus offset by the capability of the machineto satisfy its own reactive power needs and, if necessary, to delivery reactive power to the grid(it is important to realise that reactive power is not for free; industrial customers are chargedfor the consumed reactive power).

In what follows, only the synchronous machines with uniform air-gap will beelaborated. The reason is that an analysis of synchronous machines with non-uniform air-gap ismore involved, due to the existence of the reluctance torque component.

The concept of the rotating fields is reviewed next. This is followed by an explanationof the mechanism of the reactive power production in a synchronous machine. Motoring modeof operation and the generating mode of operation are then discussed in two separate sections.

2. ROTATING FIELDS AND REACTIVE POWER BALANCE OF ASYNCHRONOUS MACHINE

Consider the stator winding of a synchronous machine. Suppose that the machine

operates as a motor, supplied with a system of three phase currents( ) ( )i I t i I t i I t a m b m c m= = = cos cos / cos / 2 3 4 3 (2)

These currents are delivered by the grid and can be regarded as being impressed into the threewindings, that are spatially displaced along the circumference of the stator by 120 degrees.Current flow in each of the three phases causes an appropriate m.m.f., that acts along themagnetic axis of the given phase. The situation is illustrated in Fig. 3, where individual phasem.m.f.s are given with (N is the number of turns per phase):

( )

( )

F NI t

F NI t

F NI t

a m

b m

c m

==

=

cos

cos /

cos /

2 3

4 3

(3)

ac

b

Fa Fb Fc

bc

a

Fig. 3 - Individual phase m.m.f.s of a three-phase winding.

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E Levi, 2000 5

One observes that in terms of spatial dependence, all the three individual phase magneto-motive forces are stationary and they act along the defined magnetic axis of the winding. From(3) one notices that each of the three m.m.f.s is varying in time. The values of the three phasem.m.f.s in the given instant of time correspond to those met in any three phase system.

The resultant magneto-motive force that stems from the three phase system of currentsflowing through spatially displaced windings is the sum of the individual contributions of thethree phases. The summation is done in the cross-section of the machine, and it is necessary toobserve the spatial displacement between the three m.m.f.s. One may regard the cross-sectionof the machine as a Cartesian co-ordinate system in which phase a magnetic axis correspondsto x-axis, while y-axis is perpendicular to it. For the purposes of calculation this co-ordinatesystem may be treated as a complex plane, with x-axis corresponding to the real axis, and y-axis corresponding to the imaginary axis. In terms of the complex plane, spatial displacementof the m.m.f. by 120 degrees corresponds to the so-called vector rotator, ( )a j= exp / 2 3 .

Hence the resultant magneto-motive force can be written as

( )( ) ( )( )

F F F F e e

F NI t t t

res a b c

j j

res m

= + + = =

= + +

a a a a

a a

2

2

3 2

4

3

22 3 4 3

,

cos cos / cos /

(4)

The expression for the resultant magneto-motive force is most easily found if one recalls thewell-known correlation ( )cos . exp( ) exp( ) = + 0 5 j j . Hence

( ) ( ) ( ) ( )( )

( )

( )

F NI e e ae ae a e a e

F NI e e aa e aae a a e a a e

a a

a a a a a a a

F NI e a a aa e

res j t j t j t j t j t j t

res m j t j t j t j t j t j t

res m j t j t

= + + + + + =

+ + + + +

+ + == = = =

= + + + +

1

21

2

1 0

1

1

21 1

2 3 2 3 2 4 3 2 4 3

2 2 2 2

2

2 2 3 4

2 2

/ / / /

* *

* *

=

( )( )

( ) ( )( )

a a

F NI e e

F NI e

res m j t j t

res m

j t

2

1

23 0

3

2

+ =

= +

=

(5)

Symbol * in (5) stands for complex conjugate. The result obtained in (5) is an important one.The equation

F NI eres m j t =

3

2 (6)

describes a circular trajectory in the complex plane. This means that the resulting magneto-motive force, produced by three stationary, time-varying m.m.f.s (often called pulsatingm.m.f.s) is a time-independent and rotating magneto-motive force. Thus it follows that thethree-phase system creates a rotating field (called also Teslas field) in the machine. Figure 4

illustrates the rotating field in different instants of time. The speed at which the rotating fieldtravels equals the angular frequency of the stator three-phase supply.

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E Levi, 2000 6

Im t=90

t=135 Fres t=45

t=0Re (a)

1.5NI m

Fig. 4 - Resultant field in the three-phase machine for sinusoidal supply conditions.

Since the rotor winding carries excitation current, a field is produced by this current.This field is stationary with respect to rotor. Since the rotor rotates at synchronous speed,then, looking in from stationary stator, this rotor field rotates at synchronous speed. This isalways the case in any multi-phase AC machine: regardless of whether the rotor rotatessynchronously or asynchronously, all the fields in the machine rotate at synchronous speed.

Since the resulting m.m.f. is responsible for the resulting flux density and ultimatelyresulting flux, this means that apart from the rotating m.m.f., there is a rotating flux densitywave and a rotating flux in the machine as well. The term rotating field in general denotes anyof the three.

In other to show how a synchronous machine can produce reactive power, consider theoperation of a synchronous motor. The motor will always consume real power; however, it caneither generate or absorb reactive power. Let us further consider a couple of characteristicsituations. Let us assume at first that a synchronous motor operates under ideal no-loadconditions, so that all the losses in the motor can be neglected. This means that the input realpower and the output mechanical power are both zero. Such an ideal situation is useful inexplaining the reactive power production and consumption.

Suppose at first that the rotor current is zero. Therefore the rotor does not produce anyrotating field. Rotating field of the stator is generated due to current flow in stator and thestator rotating field equals the total rotating field of the machine (total field is a vectorial sumof the field generated by the stator currents and the rotor field; it is fixed since the supplyvoltage is fixed, and does not depend on individual values of the rotor and stator fields). Sincerotor current is zero, there is not an induced emf in the stator winding due to rotor flux. Underthese conditions the machine behaves like a pure reactance and consumes reactive power. Thesituation is illustrated in Fig. 5a, where total field and stator rotating field are shown for oneinstant in time. If the reactance of a stator phase winding is X s, then the current in one phase isI = V/X s, where V is the rms of the phase to neutral voltage. Hence the reactive powerconsumed by the machine is Q = 3 V I = 3V 2 /Xs.

Suppose next that rotor current is now increased to a certain value, say, such that theemf in the stator winding equals one half of the applied phase to neutral voltage. The situation

is illustrated in Fig. 5b. The stator field is now just one half of the value for zero rotor current,since one half of the total field is provided by rotor. The current in stator is I = (V - E) /X s, and

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E Levi, 2000 8

Ftotal Fs (= F res)

Phase a axis

a.

Ftotal Fs (= F res)

Fr

Phase a axis

b.

Ftotal Fr

Phase a axis

c.

Fs

Ftotal Fr

Phase a axis

d.

Fig. 5 - Illustration of rotating fields in a synchronous motor (for ideal no-load conditions) for four different values of the rotor current.

Fs

Ftotal Fr

Phase a axis

Fig. 6 - Rotating fields in motoring operation in over-excited mode.

jXs

I +

V E

Fig. 7 - Equivalent circuit of a synchronous motor.

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E Levi, 2000 9

The following phasor equation follows directly from Fig. 7:

E jX I s= + (7)

The phasor diagram, that this equation describes, is drawn in Fig. 8 for two cases: operation inthe under-excited mode when the power factor is lagging (since the machine consumes reactive

power) and operation in the overexcited mode when the power factor is leading (since themachine consumes active power but simultaneously delivers reactive power to the grid).

jXsI

V jX sIV E

E

I I

lagging power factor leading power factor

Fig. 8 - Phasor diagram for motoring operation in under-excited and overexcited modes.

Phasor diagrams of Fig. 8 are used most frequently to determine the unknown loadangle and the induced emf on the basis of the known motor loading and stator terminalconditions. They simultaneously represent the starting point in deriving the so-called load anglecharacteristic of a synchronous motor. Consider the phasor diagram of Fig. 8 for overexcitedmode, that is for convenience re-drawn in Fig. 9. Due to the correlation that exists, oneimmediately recognises that the angle BCA is equal to the power factor angle (angle betweencurrent and voltage phasors). On the other hand, angles ODA and ABC are right angles. Whensolving the phasor diagram of Fig. 9, it is much easier to use one of the two triangles (triangleODC or triangle OBC) and project the phasors on the sides of these two triangles, than todirectly solve the complex phasor equation (7).

In order to obtain the load angle characteristic of a synchronous motor one considersthe triangle OBC in Fig. 9. By projecting the phasors on sides OB and CB one has

CB

A jX s ID

V E

I

O

Fig. 9 - Phasor diagram for derivation of load angle characteristic.

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E Levi, 2000 10

X I E

X I E s

s

+ ==

sin cos

cos sin

(8)

When terminal voltage, stator current and the power factor are known, these two equationsenable simple calculation of the load angle and the induced emf. However, for the purpose of deriving the load angle characteristic one expresses the active stator current component fromthe second equation of (8) as

I E

X scos

sin

= (9)

Next, it is necessary to recall that the phasor diagram represents one phase of the machine andthat the machine is three-phase. Hence the input active power and the reactive power (that iseither absorbed or delivered, as in Fig. 9) are given in terms of phase to neutral voltage andphase current as

P VI Q VI = =3 3cos sin (10)

Load angle characteristic of a synchronous machine relates the active power with the loadangle, . Note that, since the stator resistance loss and the rotational losses are neglected, inputactive power equals active power transferred from stator to rotor and it ultimately equals theoutput mechanical power. Substitution of (9) into (10) yields

P VI V E

X

VE

X s s= = =3 3 3cos

sinsin

(11)

The correlation between real (or output) power and load angle is called load anglecharacteristic of a synchronous motor. One observes from (11) that, for the given statorvoltage and rotor current (that is, emf) power depends only on the sine of the load angle. Thismeans that, higher the load is, higher the value of the load angle will be, and vice versa. Inother words, rotor field in Fig. 6 will lag the total field more and more as the loading of themachine increases.

The maximum power that a synchronous motor can deliver is met when load angleequals 90 degrees. Hence, for each rotor current setting (that is, for each value of the emf)there is a maximum power that the motor can deliver. The maximum power is

PVE

X smax = 3 (12)

If the power required by the load increases above the maximum power, load angle would goover 90 degrees and the so-called loss of synchronism would take place. Since increase of theload angle above 90 degrees leads to decrease of the motor power (compared to maximumpower) rather than an increase, power demanded by the load cannot be met and the motorbecomes unstable (it looses synchronism, that is, the speed starts decreasing). The stableoperation of the motor is therefore possible only for load angle values between zero and 90degrees.

Torque of the motor can be directly obtained from the load angle characteristic bydividing the power with the mechanical synchronous angular speed of rotation:

T P

V E

X f PP

f

VE

X

T P f

VE X

es s s

es

= = =

=

31

23

1

2

3 12

sinsin

max

(13)

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E Levi, 2000 11

Hence the motor torque dependence on load angle is the same as for the power. The scalingfactor is the constant, synchronous mechanical angular speed of rotation.

Load angle characteristics of a synchronous motor are illustrated in Fig. 10, for acouple of values of the rotor current (i.e. emf). An increase in the rotor current causes anincrease in the induced emf. Consequently, for the given load, the load angle decreases. Anincrease of the rotor current leads to an increase in the maximum power (maximum torque)that the motor can develop.

Pmax1 , T emax1P, T e E1

Pmax2 , Temax2 E2

Pmax3 , Temax3 E3

STABLE UNSTABLE

E1 > E 2 > E 30 90 180

Fig. 10 - Load angle characteristic of a synchronous motor.

Reactive power characteristic of the motor can be derived in exactly the same way asthe active power characteristic. Consider the phasor diagram for under-excited operation, that

is for convenience re-drawn in Fig. 11. The following two equations now correspond to (8): X I E

X I E s

s

==

sin cos

cos sin

(14)

They are obtained by using projections of phasor in triangle ABC in Fig. 11. From (14) oneexpresses the reactive component of the stator current and then substitutes this expression intothe equation for reactive power, (10). Thus

Q VI = 3 sin (10a)

and further

I V E

X

Q VI V V E

X

V VE

X

QV

X

VE

X

s

s s

s s

sincos

sincos cos

cos

=

= =

=

=

3 3 3

3 3

2

2

(15)

The first term in this equation is the reactive power that the machine needs for its ownmagnetisation. The second term is the reactive power produced by the machine. It followsfrom (15) that reactive power is positive (i.e. absorbed) as long as V > E cos. The reactivepower is zero when V = E cos and it becomes negative (i.e. the motor delivers reactive powerto the grid) when V < E cos. Reactive power characteristic is illustrated in Fig. 12.

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E Levi, 2000 12

A jXsI

V B C

E

I

Fig. 11 - Phasor diagram of a synchronous motor for under-excited operation.

It is easy to observe in Fig. 12 that the reactive power may be both positive andnegative for the given value of the induced electro-motive force (i.e. rotor current). Whether

the reactive power is delivered or absorbed will depend on the loading of the motor, since theload at the shaft determines how much the load angle is. The internal reactive powerproduction of a synchronous motor is at maximum value when load angle is zero. Thissituation corresponds to the no-load operation, that is not normally met when the machine isused as a motor. However, this fact is made use of in so-called synchronous condensers thatare exclusively used for reactive power production and delivery to the grid. A synchronouscondenser is in essence a synchronous motor that operates at all times under no loadconditions. The load angle, although not exactly zero since there are losses in the machine (thatwere neglected throughout here) is very small, so that the machine produces maximum possibleamount of reactive power. Synchronous condensers are essentially reactive power generatorsand they are installed at various points in a power system to provide the necessary reactivepower support.

Generating mode of operation is discussed in the subsequent section. In principle, allthe characteristics remain valid, but one has to take into account that the active power will nowbe produced and delivered to the grid, rather than absorbed. This fact does cause someinevitable differences,, especially in terms of the phasor diagrams.

Q3 V(V E cos ) /Xs

3V 2 / Xs

0 90 180

3VE cos /X s

Fig. 12 - Reactive power characteristic of a synchronous motor.

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E Levi, 2000 15

V = 6300 VE =

jXs I 7.85 kV

V jX s I V E = 6.3 kV V

E = 6 kV

I I

Example 5:

Construct the phasor diagram of a synchronous motor when it operates in under-excited and over-excited mode and give the equivalent circuit.

Solution:

jXs

I +

V E

Equivalent circuit

jXsI

V jX sIV E

E

I I

lagging power factor leading power factor

Phasor diagram for motoring operation in under-excited and overexcited modes

Example 6:

A synchronous motor is connected to a 3980 V, 3-phase line and rotor current is suchthat the induced emf is 1790 V (line to neutral). The synchronous reactance is 22 Ohmsand the load angle between the voltage and the emf is 30 degrees. Determine the statorcurrent and the power factor angle.

Solution:

Since the given voltage is line to line, the corresponding phase to neutral voltage is 3980 / 3 = 2300V. As phase to neutral voltage is greater than induced emf, the motor operates in under-excited mode.The solution is then most easily found using the phasor diagram given in Figure and equations

A jXsI

V B C

E

I

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E Levi, 2000 16

X I E

X I E s

s

==

sin cos

cos sin

2300 22 1790 30 1550 185

22 1790 30 895

40 682

22 40 682 749.815

0 83777

40

0 766

40 682 531

= = = =

= =

==

== =

I

I

I

I

sin cos .

cos sin

. / cos

( . / cos ) sin

tan .

cos .

. / cos .

Hence

lagging

A

o

Example 7:

A 150 kW, 1200 rpm, 460 V, 3-phase synchronous motor has a synchronous reactanceof 0.8 Ohms and the induced emf is 300 V phase to neutral. Determine the power -load angle characteristic, torque - load angle characteristic and maximum power andtorque.

Solution:

Phase to neutral voltage is 460/ 3 = 266 V. Synchronous speed of the machine is 1200 rpm. Hence,from the table in Example 2, it follows that the machine is a 60 Hz one and that the number of polepairs is 3. Thus the power vs. load angle and the torque vs. load angle characteristics are

PVE

X

T P

f

VE

X

s

es

= =

=

= = =

3 3266 300

0 8 299.25

23 3

3

2 60299 2381 4

sin . sin sin

sin ,250sin . sin

[kW]

[Nm]

Corresponding maximum power and maximum torque are obtained, by setting load angle to 90degrees, as 299.25 kW and 2381.4 Nm.

Example 8:

A 3 MW, 6.6 kV, 60 Hz, 200 rpm synchronous motor operates at full load at a leading

power factor of 0.8. If the synchronous reactance is 11 Ohms, determine: the apparentpower that the motor develops on stator terminals, the stator current, the induced emf,and the load angle.

Solution:

As the motor operates at full load it delivers 3 MW on the shaft. All the losses are neglected, so thatthe input active power is 3 MW as well. The apparent power is then

S = 3 VI = P / cos = 3 / 0.8 = 3.75 MVAAs the line to line voltage is 6.6 kV, the corresponding phase voltage is 3815 V. The stator current isthen found from the apparent power,

S = 3 VI = 3.75 MVA I = S /(3V) = 328 A

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E Levi, 2000 17

The motor operates as over-excited (leading power factor). Hence the phasor diagram is the one shownin Figure.

CB

A jX s ID

V E

I

O

It is described with:

X I E

X I E s

s

+ ==

sin cos

cos sin

so that the unknowns are the induced emf and the load angle. Thus

( )3815 11 328 0 811 328 08

2886 4

5979.8

2886 4 5979.8 0 48272577

5979.8 2577 6640

1+ = =

==

= === =

sin cos . cos

. sin

sin .

cos

tan . / ..

/ cos .

E

E

E

E

E

o

[V]

4. GENERATING MODE OF OPERATION

In generating, active power is produced from input mechanical power and delivered to

the grid. On the other hand, reactive power may again be delivered to the grid (over-excitedmode) or it may be consumed (under-excited mode). Positive current direction in theequivalent circuit of Fig. 7 for motoring was from the grid to the machine. Hence both the realpower and the reactive power are positive when consumed. This positive direction for thegenerator stator current is inconvenient, since produced real power would come with anegative sign. It is therefore customary to change the positive stator current direction forgeneration, so that the current is positive when it flows from the machine to the grid. Thismeans that the active and reactive power are now positive when delivered to the grid (reactivepower becomes negative when consumed, i.e. for under-excited mode). The correspondingequivalent circuit is shown in Fig. 13 and it differs from the one of Fig. 7 only in the positivedirection for stator current flow. Hence the phasor equations that describes a synchronousgenerator now takes the form

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E Levi, 2000 18

E V jX I s= + (16)

This change of the positive direction of the current has some important consequences on thecorresponding phasor diagrams and definitions related to the power factor of the machine. Theissue will be clarified shortly, upon considerations of the rotating fields in the machine duringgeneration.

jXs

I +

V E

Fig. 13 - Equivalent circuit of a synchronous generator.

In motoring, as shown in Fig. 6, rotor field lags the total field by the load angle. Henceinduced emf, that corresponds to the rotor field, lags in the phasor diagram stator voltage (thatcorresponds to the total field) by the load angle. That is, the emf phasor appears to the rightwith respect to the stator voltage phasor. When the machine generates, the load angle againappears between the total field and the rotor filed. However, as illustrated in Fig. 14, rotor fieldnow leads the total field by the load angle. This means that in the corresponding phasordiagrams induced emf phasor will lead the stator voltage phasor by the load angle. In otherwords, for generation, the emf phasor will appear to the left of the stator voltage phasor.

Fs

Fr

Ftotal

Phase a axis

Fig. 14 - Rotating fields in generating operation in over-excited mode.

The other important difference with regard to the phasor diagram is related to themeaning of the lagging and leading power factor. For motoring one has the standard definitionrelated to the positive current flow into the machine: power factor is lagging when the machineconsumes reactive power (under-excited operation), while it is leading when the machinedelivers reactive power (over-excited mode). For generation however, due to the change of thepositive stator current direction, the situation reverses: over-excited mode that means deliveryof the reactive power to the grid is now described with the lagging power factor, while under-

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E Levi, 2000 19

excited mode that means consumption of the reactive power corresponds to the leading powerfactor. The phasor diagrams for the generator operation in under-excited and overexcitedmode are shown in Fig. 15. Summarising, if it is said for a generator that it operates withlagging power factor, this means that it is in the over-excited mode and delivers the reactivepower to the grid. If a generator operates with leading power factor, it is under-excited and it

consumes reactive power from the grid.

jXsI

E jXsI V

VE

I I

leading power factor lagging power factor

Fig. 15 - Phasor diagram for generating operation in under-excited and overexcited modes.

As far as the load angle characteristic of a synchronous generator is concerned, nothingchanges. Active power delivered to the grid for generation has been made positive by changingthe direction of the stator current. Hence the equations for power vs. load angle, torque vs.load angle and maximum power and torque remain to be given with (11),(12) and (13). That is,

P VI VE X s

= =3 3cos sin (11)

PVE

X smax = 3 (12)

T P f

VE

X

T P f

VE

X

es

es

=

=

31

2

31

2

sin

max

(13)

Load angle characteristic is therefore the same as in Fig. 10 and is repeated for convenience inFig. 16. Stable operating region is again for load angles between zero and 90 degrees.Remember however that the load angle represents now the angle by which the induced emf phasor leads the stator voltage phasor, since the rotor field leads the total field by the sameangle (while the angle is lagging for motoring).

As far as the reactive power characteristic is shown, reactive power is now positivewhen delivered to the grid. Hence the terms related to V and E cos in (15) interchange theposition and the equation is

QVE

X

V

X s s= 3 3

2cos (17)

The condition that describes transition from under-excited mode to over-excited mode

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E Levi, 2000 20

however remains the same. This means that the reactive power is zero when V = E cos and itbecomes positive (rather than negative, as for motor) when the generator delivers reactivepower to the grid when V < E cos. Reactive power characteristic is illustrated in Fig. 17.

Pmax1 , T emax1P, T e E1

Pmax2 , Temax2 E2

Pmax3 , Temax3 E3

STABLE UNSTABLE

E1 > E 2 > E 30 90 180

Fig. 16 - Load angle characteristic of a synchronous generator.

Q3 V(E cos V)/X s

3VE cos /X s

0 90 180

- 3V 2 / X s

Fig. 17 - Reactive power characteristic of a synchronous generator.

EXAMPLES

Example 9:

A synchronous generator operates on mains of 6.3 kV phase to neutral voltage.Synchronous reactance is 14 Ohms and the stator resistance can be neglected. Themachine operates under ideal no-load conditions. For induced electromotive forces

equal to 6000 V, 6300 V and 7850 V construct the phasor diagram, and calculatestator current and reactive power.

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E Levi, 2000 21

Solution:

Note that this example is identical to the example 4, except that the machine is given as a generatorrather than as a motor. This means that all the values calculated in example 4 will be valid, but thesigns will change.

The induced emf and the applied voltage are in all the three cases under consideration in phase since

the machine does not generate real power and all the losses are neglected. Since the voltage is 6300 Vphase to neutral, the induced emf of 6000 V means that the machine consumes reactive power (E cos = E, E < V). Stator current is

I = (E - V) / X s = (6000 - 6300) / 14 = - 21.4 A

and the consumed reactive power is

Q = 3 V I = - 3 x 6300 x 21.4 = - 404.46 kVAr

In the second case the induced emf and the applied voltage are the same, 6300 V. There isconsequently no current flow in stator and the reactive power is zero. The machine neither consumesnor delivers reactive power.

In the third case induced emf is 7850 V. It is greater than applied voltage and the machine nowdelivers reactive power to the grid. Stator current is

I = (E - V) / X s = (7850 - 6300) / 14 = 110.7 A

and the delivered (positive sign) reactive power is

Q = 3 V I = 3 x 6300 x 110.7 = 2.092 MVAr

Corresponding phasor diagrams are given in Figure. Note the change in the position of the statorcurrent phasor, compared with the phasor diagrams given in example 4. Note as well that the arrowfor the voltage drop on the stator synchronous reactance points in the opposite direction compared tothe figure in example 4, due to the opposite positive direction for current flow.

V = 6300 VE =

jXs I 7.85 kV

V jX s I V E = 6.3 kV V

E = 6 kV

I I

Example 10:

Construct the phasor diagram of a synchronous generator when it operates in under-excited and over-excited mode and give the equivalent circuit.

Solution:

Note the analogy between this example and the example 5.

jXs

I +

V E

Equivalent circuit of a synchronous generator

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E Levi, 2000 22

jXsI

E jXsI V

V

E

I I

leading power factor lagging power factor

Phasor diagram for generating operation in under-excited and overexcited modes

Example 11:

A three-phase turbo-generator has the synchronous reactance equal to 14 Ohms and

delivers to the grid real power of 1.68 MW at lagging power factor. Grid voltage is 11kV and the stator current is 100 A. Calculate the power factor, load angle and theinduced emf.

Solution:

Given voltage is line-to-line value. Phase-to-neutral voltage is 6350 V. Active power of the generatoris known, P = 1.68 MW. The apparent power is S = 3 VL-L I = 1.90525 MVA. Power factor istherefore

p.f. = cos = P/S = 1.68/1.90525 = 0.8817Power factor is lagging. Since cos = 0.8817, power factor angle is 28.15 degrees and sin = 0.4718.The generator is over excited and the phasor diagram in figure applies. Using triangle ABC one has:

A jX sI

E

VB

I

C

( )( )

( )

( )

E V

E X I V

X I V

V

E V

E E

s

s

L L

cos cos

sin sin

tansin

cos

.

..

.

. .

cos

cos

.

cos ..

+ =+ = +

+ =+

= +

=

+ == =

=+

=

=

= =

Dividing the second equation with the first equation gives:

14 100 6350 0 4718

6350 0 88170 785

38137

38137 2815 10

6350 08817

381377118 3 V

3 12.33 kV

o

o

o

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E Levi, 2000 23

Example 12:

Given a three-phase generator with rated data 16 MVA, 10.5 kV, 50 Hz, rated p.f. =0.8, two-pole, synchronous reactance = 13.77 Ohms, calculate:

a) maximum power and torque, assuming that induced emf corresponds to rated

operating conditions;b) maximum power if the generator is loaded with 10 MW, terminal voltage is only 8kV and the power factor is 0.6 lagging.

Solution:

a) Once more, given voltage is line-to-line and the phase to neutral voltage is therefore 6.06 kV. Notethat it is not specified whether for rated operation power factor is leading or lagging. However, for asynchronous machine the normal operating mode is the over-excited mode. This means that the powerfactor is lagging in generation. From the given power factor value of 0.8 one finds the power factorangle as equal to 36.87 degrees and the sin = 0.6.Since the rated apparent power is S = 3 VL-L I = 16 MVA, one finds the rated value of the statorcurrent asI = S / (3 V L-L) = 879.8 A.Phasor diagram is the same as in the previous example. Using once more the triangle ABC, one has:

A jX sI

E

VB

I

C

( ) ( )

( ) ( )

( )

E V X I V

E V V

E

P S

PVE

X

PVE

X

T P f P

s

s

s

e

= + + =

+ = + = =

+ === =

= =

=

= =

=

= =

cos sin .

cos cos coscos

.

.

cos

sin. .

.sin .

. .

..

/

max

max max

2 2 1647

0294

72.8

35 9

12.8

3 36 06 1647

137735 9 12.8

3 3606 1647

13772175 MW

2 69

kV

MW

MW

kNm

o

o

o

b) All the calculations are essentially the same, except that the values are different, including voltage.Phase to neutral voltage is now 4.618 kV. Stator current is obtained from

I = P / (3Vcos ) = 1200 Aand one concludes that the machine is overloaded (rated current is 879.8 A). Since power factor is 0.6,then power factor angle is 53.13 degrees and sin = 0.8. Further

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E Levi, 2000 24

( ) ( ) E V X I V

PVE

X

PVE

X

s

s

s

= + + =

= = =

=

= =

cos sin .

sin sin .

.max

2 2 20 4

3 10 0 487

29.15

3 20 52

kV

MW

MW

o

5. TUTORIAL QUESTIONS

Q1. A three-phase two-pole synchronous generator with cylindrical rotor structure has therated data 11 kV, 50 Hz, 20 MVA, star connected stator winding. Synchronousreactance is 5 and rated power factor is 0.9 lagging.a) Determine value of induced electromotive force and load angle for rated operatingconditions. Find the maximum real power that the generator may deliver to the grid forthe calculated induced electromotive force.b) Determine induced electromotive force and load angle if the generator now operateswith power factor of 0.8 lagging, delivers to the grid 12 MW and the grid voltage is 10kV.c) Sketch the two power - load angle characteristics and denote all the calculatedpoints.

Q2. Two identical three-phase star-connected synchronous generators operate at 33 kV andeach provide 5 MW for the load of 10 MW, 0.8 lagging power factor. The synchronousreactance of each machine is 6 . The first machine operates with 125 A current atlagging power factor.a) Calculate the current and power factor of the second machine.b) Determine load angle and induced emf for both machines.

Q3. A three-phase two-pole synchronous generator with cylindrical rotor structure has therated data 11 kV, 50 Hz, 20 MVA, star connected stator winding. Synchronousreactance is 5 and rated power factor is 0.9 lagging.a) Determine values of induced electromotive force and load angle for rated operatingconditions. Find the maximum real power that the generator may deliver to the grid for

the calculated induced electromotive force.b) Determine induced electromotive force and load angle if the generator now operateswith power factor of 0.7 lagging and delivers to the grid 12 MW at the rated gridvoltage.c) Sketch the two power - load angle characteristics and denote all the calculatedpoints.

Q4. A synchronous machine has a three-phase stator winding, excitation rotor winding andcylindrical structure of both stator and rotor.(a) Sketch equivalent circuit and appropriate phasor diagram if the machineoperates as:

(i) overexcited generator (ii) underexcited generator(iii) overexcited motor (iv) underexcited motor

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Define for each case direction of real power and reactive power flow between machineand grid and specify the power factor (leading or lagging).b) A three-phase two-pole synchronous generator with cylindrical rotor structure hasthe rated data 16 MVA, 10.5 kV, 50 Hz, star connected stator winding. Synchronousreactance equals 13.77 and rated power factor is 0.8 lagging. Determine values of induced electromotive force and load angle for rated operating conditions and calculatemaximum real power that the generator may deliver to the grid, if the inducedelectromotive force is the same as for rated operating conditions.

Q5. a) A three-phase turbogenerator has synchronous reactance equal to 14 ohms anddelivers to the grid real power equal to 1.68 MW at lagging power factor. Thegenerator has star connected stator winding, operates on 11 kV grid and the statorcurrent is 100 A. Calculate the power factor, load angle and induced electromotiveforce.b) A synchronous and an induction generator are connected in parallel to the same

busbars and they deliver together real power of 800 kW with overall power factor of 0.8 lagging. Induction generator operates with power factor of 0.9 and delivers realpower equal to 300 kW to the grid. Determine the real power delivered by thesynchronous generator and the power factor of the synchronous generator.

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