SYNCHRONOUS MACHINES

HARSHITA SHARMADepartment of Electronics and Communication EngineeringCommunication Engineering

JIIT, Sec‐128, Noida

INTRODUCTIONINTRODUCTION• They are synchronized as rotor speed is exactly fixed by the supply

frequency.• The other type of ac machines are called asynchronous (or Induction).

Wh h i d b i i k l• When the rotor is rotated by a prime‐mover, it works as an alternator(Synchronous generator).

• To run it as a synchronous motor:y– Three‐phase supply is connected to the stator winding

(Armature), andA d l (f th fi ld) t th t i di t d– A dc supply (for the field) to the rotor winding to producemagnetic poles.

• Important Parts of the Synchronous Machines are:Important Parts of the Synchronous Machines are:– Stator (Armature windings)– Rotor (Field windings)( g )– Damper windings (To prevent hunting in Generator and to provide

starting torque in Motors).

• General Characteristics of the SynchronousMachine:– The dc field circuit (on the rotor), is a rotatingelectromagnet controlled by the dc exciting current.

– The power supplied to the dc field circuit does notp ppenter directly into the energy‐conversion process.

– The armature circuit is placed on the stator andpcarries three‐phase currents.

– The flow of real power through the system isp g ydetermined by the mechanical input because themechanical system exchanges real power only.

– The reactive power flow is controlled by the dc fieldcurrent. It is a surprising and extremely usefulproperty of the synchronous machines.

• Synchronous speed:Synchronous speed:– One revolution of rotor represents 360 mechanicaldegrees.degrees.

– One cycle of emf represents 360 electrical degrees.– For 2‐pole machine, mechanical and electrical degrees areFor 2 pole machine, mechanical and electrical degrees areidentical.

– If the machine has 4 poles, one cycle of emf would bep , ygenerated when the field structure (on the rotor) rotatesthrough one‐half revolution only.

– Thus, in a 4‐pole machine two cycles of emf is generatedwhen the rotor completes one revolution.

– If the machine has P poles, the number of cycles of emf inone revolution will be P/2.

In general the electrical angle θ and mechanical– In general, the electrical angle θe and mechanicalangle θm in a machine are related as:

Pme

P θθ )2

(=

– If rotor has speed ns revolutions per second,frequency f of induced emf is:q y

snPf )2

(=

– The synchronous speed Ns (expressed inl ti i t ) i irevolutions per minute) is given as:

fN 120=

PNs

• What can be the maximum speed of asynchronous machine in India ?yAns. 3000 rpm. A machine cannot have lessthan 2 polesthan 2 poles.

• What can be the maximum speed of asynchronous machine in USA ?synchronous machine in USA ?Ans. 3600 rpm.

Example 1: A six‐pole ac generator is runningand producing voltage at a frequency of 60 Hz.Calculate the revolutions per minute of thepgenerator. If the frequency of the generatedvoltage is required to be decreased to 20 Hzvoltage is required to be decreased to 20 Hz,how many poles would be needed on the

t if it till t th d ?generator, if it still runs at the same speed ?

Solution: 120 120 60fN ×= = = 1200 rpm

6sNP

1200 rpm

120 120 20fP × 2120 120 201200

fPN

= = = 2

SYNCHRONOUS GENERATOR (ALTERNATOR)

• The most commonly used machine for thegeneration of electric power is theg pSynchronous generator called an alternator,as it generates acas it generates ac.

• The synchronous generator is the workhorsef h lof the electric power industry.

• The armature winding is placed on the statorThe armature winding is placed on the statorand the field poles on the rotor.

• Construction:1. Stator:. Stator:– Core made of CRGO (Cold Rolled Grain Oriented) sheetsteel or silicon steel.

– Laminations in core to minimize eddy‐current losses.– Laminations are stamped out in complete rings or inp p gsegments.

– Stampings have uniformly distributed open or semi‐closedslots to accommodate armature conductors.

– Whole structure held in cast iron frame.– From an electrical viewpoint, stator of synchronousmachine and induction machine are identical.

– The windings are usually connected in y‐configuration.– Voltage per phase is 1/ or 58% of line voltage, permitting

d i i h f di l i i l i3

a reduction in the amount of dielectric insulation.

2. Rotor:– Cylindrical structure which can rotate inside statorleaving a very small air gap.

– Consists of windings to produce dc magnetic field.– Windings are excited by a separate dc generator calledWindings are excited by a separate dc generator calledExciter.

– Exciting current supplied to rotor windings throughExciting current supplied to rotor windings throughtwo slip rings and carbon brushes.

– Depending on type of prime mover used to drive theDepending on type of prime mover used to drive thealternator, there are two types of rotors:(i) Cylindrical/Non‐Salient type (slow speed)(i) Cylindrical/Non‐Salient type (slow speed)(ii)Salient or Projected‐pole type (high speed)

– Three methods of field excitation:(i) Slip rings link the rotor’s field windings to external dcsource.

(ii) DC i A d i b il f h f(ii) DC generator exciter: A dc generator is built of same shaftas the ac generator’s rotor. A commutator rectifies thecurrent sent to the field winding.g

(iii) Brushless exciter: An ac generator with fixed fieldwindings and a rotor with a three phase circuit. Diode/ SCRrectification supplies dc current to field winding.

3. Damper Winding:– In addition to dc field windings a squirrel cage winding is also addedIn addition to dc field windings, a squirrel cage winding is also added.– Under normal conditions, the winding doesnot carry any current.– When sharp changes in the loading occurs, the rotor speed begins to

fluctuate, producing momentary speed variations.– Hence, the rotor starts oscillating. The oscillation is called hunting.

During hunting there is a relative motion between rotor (damper– During hunting, there is a relative motion between rotor (damperwinding) and synchronously rotating magnetic field. So large currentflows in damper winding and emf induces as per Faraday's law as long

h l i i ias the relative motion exists.– So induced torque in damper winding acts in opposition to

instantaneous relative motion.

ROTATING MAGNETIC FLUX DUE TOROTATING MAGNETIC FLUX DUE TO THREE‐PHASE CURRENTS 

• Consider a cylindrical magnetic structure withone winding excited by a single‐phase current.

• If the current i is dc the flux density in the air• If the current i is dc, the flux density in the airgap will also be dc and it will have a maximumvalue along the horizontal plane. (Bm=Ki)

• Flux density is a function of angle θ given as:Flux density is a function of angle θm given as:cosm mB B θ=

• If the current is alternating, so will be the flux density.h f h fl d l h f• Therefore, the flux density along the air gap is functionof time t as well as angle θm,

( )B Bθ θ• Next, consider a two‐pole magnetic structure woundwith three coils separated by 120° in space

( , ) cos cosm m mB t B tθ ω θ=

with three coils separated by 120 in space.• These coils are supplied three‐phase (R Y B) currents.

Due to three‐phase currents in three phase structurethree‐phase structure 

• The current iR enters into the bottom conductors andRreturns from the top.

• The current iY enters in the first quadrant and current iBY Benters in the second quadrant.

• The coil mmfs are tapered sinusoidally.• The maximum flux density from coil R lies in thehorizontal plane.

• The maxima from coils Y and B are displaced 120° and240° in space, respectively.

• Analysis of Three‐Phase System:– The three‐phase currents supplied,

( ) cos ( )R mi t I tω=( ) cos ( 120 )Y mi t I tω= − °( ) cos ( 240 ) cos ( 120 )B m mi t I t I tω ω= − ° = + °

– The flux density from the three coils,y ,

( , ) cos( 120 ) cos( 120 )YB t B tθ ω θ= − ° − °

( , ) cos( ) cos( )R m m mB t B tθ ω θ=

( , ) cos( 120 ) cos( 120 )Y m m mB t B tθ ω θ

( , ) cos( 120 ) cos( 120 )B m m mB t B tθ ω θ= + ° + °

( , ) ( , ) ( , ) ( , )cos( ) cos( ) cos( 120 ) cos( 120 )

cos( 120 ) cos( 120 )

m R m Y m B m

m m m m

B t B t B t B tB t B t

B t

θ θ θ θω θ ω θ

ω θ

∴ = + += + − ° − °

+ + ° + °cos( 120 ) cos( 120 )m mB tω θ+ + ° + °

1

2( , ) {cos( ) cos( )}m m m mB t B t tθ ω θ ω θ= − + +

1

2

1

{cos( ) cos( 240 )}

{cos( ) cos( 240 )}

m m mB t t

B t t

ω θ ω θ

ω θ ω θ

+ − + + − °

+ + + + °2

{cos( ) cos( 240 )}m m mB t tω θ ω θ+ − + + +

3

2cos( )mt

Bω θ⎡ ⎤−

⎢ ⎥=1

2{cos( ) cos( 240 ) cos( 240 )}

m

m m m

Bt t tω θ ω θ ω θ

⎢ ⎥=⎢ ⎥+ + + + − ° + + + °⎢ ⎥⎣ ⎦

– The three bracketed terms add to zero at all times.– Therefore, 

3( , ) cos ( ) cos ( )m m m r mB t B t B tθ ω θ ω θ= − = −

Where:

2( , ) ( ) ( )m m m r m

(3 / 2)B B

• Properties of the Flux Density :

(3 / 2)r mB B=

Properties of the Flux Density :1. The magnitude is 50 % greater.

2 If i i fi d h fl i i id l i ( i )2. If time is fixed, the flux is sinusoidal in space (air gap)with the maximum flux density at θmax = ωt, and if t = 0the maximum flux occurs at θm = 0°.m

3. At a fixed θm, the flux density magnitude is sinusoidal intime.

ARMATURE WINDING• Advantages of having armature winding on the

t t d fi ld i di th R t :stator and field winding on the Rotor:– It is easier to provide insulation to armature winding forhi h lt th t ti i di i t bj t d thigh voltages, as the stationary winding is not subjected tomechanical stress due to centrifugal forces and also morespace is availablespace is available.

– The external three‐phase circuit can directly be connectedwith fixed terminals on the stator, without the need of slip‐with fixed terminals on the stator, without the need of sliprings.

– For dc supply to the rotor field winding, only two slip‐rings,pp y g, y p g ,each capable of handling much smaller current andrequiring insulation for much lower voltages, are needed.

– The revolving field system is light in weight, and thereforecan run with high speed.

• In practice, the coils are short pitched and thep , pwinding is distributed.

• Hence the rms value of the induced emf is• Hence, the rms value of the induced emf isreduced by the pitch factor kp and distributionfactor k to give:factor kd, to give:

(2 ) 1 11 2 4 44d dE f T k k f Tk k= Φ × × × × = Φ

• Pitch factor or coil span factor (kp ):

(2 ) 1.11 2 4.44p d p dE f T k k f Tk kΦ × × × × Φ

Pitch factor or coil span factor (kp ):– In a full pitch coil AB, the EMFs induced in the twocoil sides Ea and Eb are in phase and resultant EMFcoil sides Ea and Eb are in phase and resultant EMFis:

Er=Ea+Eb=E+E=2EEr Ea Eb E E 2E

– In a short pitched coil, EMFs Ea and Eb induced are out of phase by angle β.The resultant EMF is given as:

β/ β/Er=OQ=2OS=2OPcosβ/2= 2E cosβ/2– The factor by which emf per coil is reduced because of pitch being less than

full pitch is called pitch factor kp.p p p

– It is given by:p

phasor sum of the coil-side emfs cos ( / 2)arithmetic sum of the coil-side emfs

h i th l b hi h il h t it h d

k β

β

= =

– For nth harmonic, it is given as:

where, is the angle by which coilsare short-pitched.β

)2/cos( βnkpn =

• Distribution factor or breadth factor (kd ):

– If q (number of slots per pole per phase) is very large, the angle α becomes very small, then  

2/i2/i

Th t t l l i ll d h d

2/2/sin

2/2/sin

σσ

αα

==q

qKd

– The total angle qα is called phase spread.– Where, σ=phase spread=600 electrical for 3‐phase and 900electrical for 2‐phase machineselectrical for 2 phase machines.

– For nth harmonic,2/

2/sinσσ

nnKdn =

Example 1:p

E l 2 A 3 h 50 H 20 l liExample 2: A 3‐phase, 50‐Hz, 20‐pole, salient‐pole alternator with star‐connected statorwinding has 180 slots on the stator. There are8 conductors per slot and the coils are full‐8 conductors per slot and the coils are fullpitch. The flux per pole is 25 mWb. Assumingsinusoidally distributed flux calculatesinusoidally distributed flux, calculate(a) the speed,(b) the generated emf per phase, and(c) the line emf(c) the line emf.

Solution:Solution:Total number of armature conductors,  180 8 1440Z = × =

Therefore, the number of turns per phase, 1440 / 2 240

3T = =

120 120 50f(a) The speed,  s

120 120 5020

fNP

×= = = 300rpm

(b) Since the coils are full pitch the pitch factor k = 1 Now(b) Since the coils are full‐pitch, the pitch factor, kp = 1. Now,180No. of slots per pole 920

= =0

E lectrical angle per po le 180S lo t angle, 20N um ber of slo ts per pole 9

α °∴ = = = °

p p

Number of slots per pole per phase,  9 33

q = =3

q

dsin ( / 2) sin (3 20 / 2)

i ( / 2) 3 i (20 / 2)qk α × °

∴ = = =°

0.960

Th f h l f h d f h

d sin ( / 2) 3 sin (20 / 2)q α × °

Therefore, the rms value of the generated emf per phase,

p d4.44E f Tk k= Φ

4.44 50 0.025 240 1 0.960= × × × × × = 1278.7 V

(c) Since, the stator winding is star‐connected, the line emf,

3 1278 7E = × = 2214 8 VL 3 1278.7E = × = 2214.8 V

ARMATURE REACTIONARMATURE REACTION

• The generator is loaded , Ia flows in the armature winding.• The load current produces a rotating flux,Φa due to mmf Fa.• Fa is called armature reaction, which rotates at synchronous

speed and in the direction of the rotor.• Thus the resultant mmf will be Phasor sum Fr=Ff + Fa• This rotating flux Фa induces a ac three phase voltage in the

stator winding.• This voltage is‐

– subtracted from the induced voltage.– represented by a voltage drop on the synchronous reactance.

• Thus the equivalent circuit of a synchronous generator is avoltage source and a reactance connected in series.

• Nature of armature reaction: – Assuming that the armature resistance andleakage reactance are negligible so that:leakage reactance are negligible so that:

V t = E r due to Φ r

– Case 1: When current and generated EMF are inphase:

– Case 2: When current lags generated EMF by 900:

– Case 3: When current leads generated EMF by 900:

• In motoring action cases will exactly reversed in all• In motoring action cases will exactly reversed in allthe three cases as the direction of Ia is reversed.

EQUIVALENT CIRCUIT OF ALTERNATOR

• E: Excitation emf induced in the stator winding due to fluxd d b it ti t i fi ld i di tproduced by excitation current in field winding on rotor.

• Xs: Synchronous reactance per phase• R: Resistance per phase• V: Terminal voltage• X1: Leakage reactance but X1<< Xs hence, ignored.• Hence for alternator, equivalent circuit (for one phase):

• Synchronous impedance Z :Synchronous impedance Zs:

Ind ced EMF E

0122 90)/(tan ∠≈∠+=+= −sssss XRXXRjXRZ

• Induced EMF E:)( ssz jXRIVIZVVVE ++=+=+=

• Rotor power angle (δR):– Phase angle between excitation EMF E and generatedvoltage V.

– Also called power angle or torque angle.– In generating action : the field poles are driven aheadof the resultant field by an angle δR by the primemovermover.

– In case of motor : the field poles lag behind theresultant field by an angle δ due to the loadresultant field by an angle δR due to the load.

• Phasor diagrams:g– Inductive load (lagging pf):

– Capacitive load (leading pf):

VOLTAGE REGULATIONVOLTAGE REGULATION

i h h i h i l l• It is the change in the terminal voltagebetween no‐load and full load expressed asper‐unit value or percentage of full‐loadvoltage.

• Per‐unit voltage regulation:V

VE −

• Percentage voltage regulation: %100×−V

VE

• V is terminal voltage at full load and E isterminal voltage when load is removed.

Example 3:A three‐phase, 600‐MVA alternatorhas a rated terminal voltage of 22 kV (line).The stator winding is star‐connected and has aresistance of 0.014 Ω per phase and asynchronous impedance of 0.16 Ω per phase.y p p pCalculate the voltage regulation for a loadhaving a power factor of 0.8 lagging.g p gg g

Solution: The full load current for the statorSolution: The full‐load current for the statorwinding is the same irrespective of the powerf t f th l d d i i bfactor of the load and is given by:

66 3

L h L600 10600 10 3 (22 10 ) 15.7 kAI I I ×

× = × ⇒ = = =L ph L 3600 10 3 (22 10 ) 15.7 kA

3 22 10I I I× × ⇒

× ×

The terminal voltage per phase on full load, 22 kV 12.7 kV;

3V = =

3z s (15.7 kA) (0.16 ) 2.512 kVV IZ= = × Ω =

0 014OC 10.014cos 0.0875 and cos 0.0875 84.980.16

OCOG

θ θ −= = = = = °

For pf = 0 8 lagging :For pf = 0.8 lagging :

1cos 0.8 36.87φ −= = °

( ) 84.98 36.87 48.11 and cos 0.6677α θ φ α= − = ° − ° = ° =

Putting all the voltages in kV we getPutting all the voltages in kV, we get 

2 2z z2 cosE V V V V α= + + ⋅ ⋅2 2(12.7) (2.512) 2 12.7 2.512 0.6677 14.5 kV= + + × × × =

The voltage regulation is

14.5 12.7 0.1417 per unit12.7

E VV− −

= = = 14.17 per cent

MEASUREMENT OF SYNCHRONOUS IMPEDANCEMEASUREMENT OF SYNCHRONOUS IMPEDANCE

• The synchronous impedance Zs of analternator can be determined by plotting itsy p gopen‐circuit and short‐circuit characteristics.

• Open‐Circuit characteristics:Open Circuit characteristics:

• Short‐Circuit characteristics:

• Synchronous impedance:Synchronous impedance:

Zpu=ZΩ/ZB

Example 4:

ALTERNATOR CHARACTERISTICSALTERNATOR CHARACTERISTICS• Terminal voltage Vs field current: TheTerminal voltage Vs field current: Thefield current equivalent to OF is required tocompensate F and F i e armature reactioncompensate Fa and Fal i.e. armature reactionmmf and leakage flux equivalent mmf .

O F = F a + F a l

• Terminal Voltage Vs Load current (Effect of• Terminal Voltage Vs Load current (Effect ofarmature reaction on terminal voltage):

Vt = E– IR – jIXs

SYNCHRONISATION OF GENERATORSYNCHRONISATION OF GENERATOR

POWER DELIVERED BY ALTERNATORPOWER DELIVERED BY ALTERNATOR• Operation in large system ‐ Infinite bus:

– The power system is modelled as an infinite bus whichmaintains constant frequency and constant voltage.

– Here, if we increase the mechanical drive, we do notincrease the frequency as in stand‐alone system; rather,we contribute larger real power to the gridwe contribute larger real power to the grid.

– Likewise, if we increase the dc field current, we do notincrease the output voltage as in stand alone system;increase the output voltage as in stand‐alone system;rather, we change the reactive power contributed to thesystem.y

• If we increase the mechanical drive to theIf we increase the mechanical drive to thealternator, the rotor power angle δR increasesand as a result the real power, delivered to thea d as a esu t t e ea po e , de e ed to t einfinite bus increases.

• The reactive power delivered by the alternatorThe reactive power delivered by the alternatorcan be controlled by controlling the dc excitingcurrent If.current If.

• If we increase If, the magnitude of excitationvoltage E increases If we keep the mechanicalvoltage E increases. If we keep the mechanicaldrive constant (i.e., the angle δR unaltered), threeconditions by just varying the dc excitationconditions by just varying the dc excitationcurrent If.

SYNCHRONOUS MOTORSSYNCHRONOUS MOTORS

i i diff f d i ?• How it is different from an Induction Motor ?– A synchronous motor always runs at synchronousspeed, whereas an induction motor runs at aspeed slightly less than the synchronous speed.Th it i t t d t i tThus, it is constant speed motor in true sense.

– In synchronous motor, there are magnetic polesth t I d ti t h tion the rotor. Induction motor has no magnetic

poles on the rotor instead it has short circuitedwindingswindings.

– Induction motors are self starting but synchronousmotors are not self startingmotors are not self starting.

• Equivalent Circuit of synchronous motor:q y

• The starting torque of synchronous motor is zero:– This is because the rotor poles are stationary with N & S‐poles– This is because, the rotor poles are stationary with N & S‐poles

on it. Stator poles with N & S poles on it are rotating withsynchronous speed.

– When the N pole of stator comes near to S pole of the rotor– When the N‐pole of stator comes near to S‐pole of the rotor,rotor experiences the maximum torque and when the S‐pole ofthe stator comes to S‐pole of the rotor, experiences themaximum torque in the opposite direction due to repulsion.maximum torque in the opposite direction due to repulsion.Thus , the average value of the torque is zero.

– When a 3Ø balanced voltage is applied to a 3Ø balancedwinding of synchronous motor a rotating magnetic field withwinding of synchronous motor, a rotating magnetic field withsynchronous speed is produced.

– If rotor is stationary (at starting), then stator field is equal to astator field rotating at a synchronous speed w r to the rotorstator field rotating at a synchronous speed w. r. to the rotorfield. Thus the angle between the two fields λ= ωst is afunction of time.

– the average value of this torque is Zero as the average value of– the average value of this torque is Zero, as the average value ofsinωst is zero.

• Starting a synchronous motor:

(a) Starting by Using Damper Winding

(b) Starting as an Induction Motor by short circuiting thet i di d l d h it i l t drotor winding dc supply and when it is accelerated near

to synchronous speed dc supply is provided(c) Starting By Using small DC Motor (Pony Motors)(c) Starting By Using small DC Motor (Pony Motors)

OPERATION OF SYNCHRONOUS MOTOROPERATION OF SYNCHRONOUS MOTOR

• Phasor Diagrams for a Synchronous Motorg ydriving a constant load for:(a) Under excitation (b) Normal excitation(a) Under excitation (b) Normal excitation(c) Over excitation

(c) (a)

(b)

• Effect of Excitation on Power Factor andArmature Current(V‐Curves):

SYNCHRONOUS CONDENSERSYNCHRONOUS CONDENSER

•An overexcited synchronous motor is used as a phase modifier or compensator.

• Employing synchronous condenser to correct• Employing synchronous condenser to correct the load power factor :