Synchronous Generator

Chapter No. 06

Synchronous Generator

Dr. Intesar Ahmed, Engr. Kashif Imran,

Engr. Muhammad Shuja Khan

Department of Electrical Engineering

COMSATS Institute of Information Technology

Lahore

Introduction

In power stations, 3-Φ AC generator is used to produced 3-Φ voltage with

low magnitude. This low magnitude of voltage is increased by using power

transformer.

The machine which generates 3-Φ electric power from the mechanical power

Is called an alternator or synchronous generator. The shaft of the generator is

coupled with mechanical turbine. The turbine may be gas, steam or water.

A synchronous machine with no-load is called asynchronous condenser or

dynamic capacitor (used to correct the power factor).

ConstructionStator

The stator is the stationary part and is made of thin laminations of highly

permeable steel to reduce losses.

The stator has ample number of slots in its inner periphery for stator winding.

The conductors are placed in these slots in a suitable way. The ends of the

conductors are connected either in Υ or Δ to form the balanced 3-Φ winding.

Rotor

It is the inner part of the alternator. It contains the field windings which are

energized by direct current from the separate dc source.

Types of AlternatorSalient Pole Type

This generators have many poles (50). Therefore, salient pole rotor is used inlow and medium speed alternators.

This alternator is driven by diesel engine or water turbine.

Salient poles are fixed to the shaft of the alternator by bolts.

The air-gap b/w stator & the rotor is not uniform.

Non-salient or cylindrical pole Alternator

The rotor is made of solid steel and the air-gap is uniform.

This rotor is used in the large generator with two or four poles.

Steam turbine is used to drive this type of an alternator.

Normally runs at high speed (3600rpm or 1800rpm).

Pole and Frequency

The number of poles of the alternator is normally depends on the speed. The frequency is directly proportional to the speed of the rotor. to the number of pairs of poles. In general,

ƒ = PN/120

Working Principle Rotor winding is initially energized by a separate dc source.

A dc field current flows through the rotor field winding that establishes a fluxin the air-gap which does not vary with time.

A rotating field of constant magnitude is produced in the air-gap when therotor rotates due to a prime mover.

The stator conductors are cut by this rotating flux, so that a voltage is inducedin the armature conductors. If the rotor rotates continuously, then 3-Φ ACvoltage will be induced in the armature conductors.

These induced voltages have the same magnitudes but phase-shifted by120°electrical. Consider the stator conductors of the alternator are Υ-connected &the rotor is energized by a dc supply.

Coil and Winding Element

With reference to fig, the two conductors AB and CD along with their end

connections constitute one coil of the armature winding. The coil may be

single turn or multi-turn coil. A single turn coil will have two conductors. But a

multi-turn coil may have many conductors per coil side.

Full Pitch and Short Pitch Winding

The coils are arranged according to the number of slots/pole/phase in the

alternator. The two sides of a coil are placed in the slots with a span of 180°.

The span between two adjacent opposite poles is known as pole pitch (זp).

The winding consisting of coils having span,

which is equal to one pole pitch i.e. spanning over

180 electrical degrees, is known as full pitch winding.

The winding consisting of coils having span, which is less

than the full-pitch winding is called the short pitch winding.

S is the total number of slots, P is the total number of Poles, m is the number

of phases, n is the number of slots per pole per phase = S/P*m

“α” angle between two successive slots = (360/S)/(P/2)=(180P/S)° elect.

pז is the pole pitch =S/P=180° elect.

cז is the coil pitch,

cז =pז for full pitch coil and זp< זc for short pitch

ץ is the short pitch angle

ץ -pז = cז °(cז-180) = electrical.

Pitch or Chording Factor The pitch factor is defined as the ratio of voltages generated by the short pitch

winding to that which is generated in the full pitch winding.Kp= (voltage induced in the short pitch coil)/(voltage induced in the full pitch coil)

Let the rms value of the coil side voltage is Ecs and the resultant coil voltage is Ecr.

For full pitch winding: Ecr=2Ecs

For short pitch winding: Ecr=2Ecs cos( 2/ץ)

Finally the pitch factor becomes, kp=cos( 2/ץ)

Practice Problems

The number of slots of a 3-Φ, 8 pole alternator is 144. Determine the pitch

factor if the winding is shorted by two slot-pitches. The number of slots of a

3-Φ, 4 pole alternator is 40.Determine the pitch factor if the coil spans from 1

to 9. The number of slots of a 3-Φ, 4pole alternator is 36. Find the pitch

factor if the winding is shorted by one slot-pitch and the coil span from slot

to 8.

Distribution Factor It is defined as the ratio of the phasor sum of the voltages in each coil to the

arithmetic sum of the voltages.

The total induced voltage in winding will be equal to sum of the induced voltage in each coil.

The voltages in each coil are not in phase but displaced by some angles due to different position of the slots. Mathematically it can be expressed a

kd = Phasor sum of coil voltages/Arithmetic sum of coil voltages.

Consider n numbers of coils are connected in series to form a phase group and the phase angle between the slots is α.

The distribution factor is, kd=sin (nα/2)/(nsin α/2)

Practice problem A 4-pole alternator has 48 slots where double layer windings are placed. Find

the distribution factor.

An 8-pole, 3-Φ alternator has 144 slots for double layer winding. Find the distribution factor.

Effect of Harmonics on Different Factors

The induced voltage in the armature windings will contain harmonics due to

non sinusoidal space flux density distribution. These harmonics also affect

the pitch and the distribution factor. Consider the ץ is the short pitch angle for

the fundamental flux wave.

The pitch factor is, kp = cos ץ

The distribution factor is,kd = (sin nα)/ (nsin α/2)

Then the pitch and distribution factor for hth harmonic are,

kph = cosh2/ץ

kdh = sin(nhα/2)/(nsin(hα/2))

EMF Equation of an Alternator

The flux created by the rotor is revolving is passing through the air-gap

with synchronous speed. This flux varies sinusoidally.

Φ=Φm sinωtAccording to Faraday’s law, the voltage induced in a single turn of coil is,

e=dΦ/dt → e=d/dt (Φmsinωt) → e=Φmωcosωt,

e=Φmωsin(90°-ωt)

The rms value of induced voltage in a single turn of coil,

E = Φmω/√2, for T turns, E = 4.44fΦmT

The general expression of emf /phase can be obtained incorporating the

pitch factor and distribution factor.

E= 4.44 kpkdf Φm T, E= 4.44 kwf ΦmT

T is the number of turns in any phase, kw = kpkd is the winding factor and Φm

is maximum value of flux

Practice Problems

The number slots of a 3-phase, 12-pole, Y-connected alternator has 180.The

number of conductor per slot is 12 and the conductors are connected in

series. The coil is full-pitch winding and the flux per pole is 0.05Wb. If the

machine runs at a speed of 600rpm, determine the line voltage.

A 3-phase,50Hz, 4-pole, Y-connected alternator has 6 slots per phase. The

armature has double layers winding. The windings are arranged as 5

conductors per layer. The coil pitch is 5/6th of the full pitch and the flux per

pole is 0.05Wb. If the machine runs at a speed of 600rpm, determine the

phase voltage.

A 3-Φ, 50Hz, 10-pole, synchronous generator has 3 slots/phase. The

armature has double layers winding. The windings are arranged as 5

conductors/layer. The coil span is 150°. The value of the fundamental and 3rd

harmonic component of the flux is 0.05wb and 0.006wb. Determine the rms

value of the induced voltage per phase voltage.

The number slots of a 3-Φ, 4-pole, Υ-connected alternator has 36 slots. The

number of conductor per slot is 8 and the conductors are connected in

series. The coil is full pitch winding and the flux per pole is 0.06Wb. If the

machine runs at a speed of 600rpm, determine the line voltage.

The number slots/pole/phase of a 3-Φ, 50Hz, 4-pole, Y-connected alternator

is 4. The armature has double layers winding. The windings are arranged as

4 conductors/layer. The coil pitch is 3/4th of the full pitch and the flux/pole is

0.03Wb. Determine the phase and line voltage.

A 3-Φ, 60Hz, 6-pole, synchronous generator has 2 slots/ phase. The

armature has double layers winding. The windings are arranged as 4

conductors/layer. The coil span is 150°. The value of the fundamental of the

flux is 0.05Wb and third harmonic component of the flux is 30 percent of the

fundamental flux. Determine the rms value of the induced voltage/phase

voltage.

Equivalent circuit of a Synchronous Generator

The stator current will flow, if the stator terminals are connected to 3-Φ load. This current will establish the rotating field in the air gap.

The resultant air gap flux is the sum of the field flux and the armature reaction flux.

These fluxes induce the field voltage Ef and the armature reaction voltage Ear. The induced voltages lag their respective fluxes by 90° electrical.

The phasor diagram is drawn including these voltages is, shown below

In the equivalent circuit the following parameters are defined as,

Ef is the induced emf/phase, Vt is the terminal voltage/phase

Ia is the armature current/phase, Xℓ is the leakage reactance/phase

Xs is the synchronous reactance/phase

Xar is the armature reaction reactance/phase

The sum of the armature leakage reactance and armature reaction reactance

is known as the synchronous reactance, Xs=Xl+Xar

The synchronous impedance of the alternator, Zs=Ra+jXs

The induced emf/phase is, Ef =Vt+IaRa+jIaXs

The magnitude of induced emf/ phase, Ef =(Vt+IaRa)2+(IaXs)2

Phasor Diagrams and Voltage RegulationIn generator three types of phasor diagrams are considered. These are unity,

lagging, and leading pf. In the entire phasor diagram, Vt is considered as the

reference phasor.

For Unity pf load

The Ia in phase with Vt. The resistive drop IaRa in phase with Ia.

The reactive drop IaXs leads the IaRa by an angle of 90°.

The induced emf /phase is, Ef=√(Vt+IaRa)2+(IaXs)2

Lagging pf factor

The Ia lags behind the Vt by an angle of φ.

The IaRa in phase with the Ia.

Reactive drop leads the resistive drop by 90°.

The induced emf /phase is,

Ef = √(Vtcosφ+IaRa)2+(Vtsinφ+IaXs)2

Leading pf

Ia leads the Vt by an angle of φ and Ra in phase with the Ia.

Xs leads the resistive drop by an angle of 90°.

The induced emf /phase is,

Ef=√(Vtcosφ+IaRa)2+(Vtsinφ-IaXs)2

Voltage Regulation

It is the change in magnitude of the terminal voltage b/w no-load and full-load

conditions.

VR=(Vnℓ-Vfℓ)/Vfℓx100=(Ef-Vt)/Vfℓx100

Practice Problems

A 50kVA, 220V, 50Hz, 1-Φ alternator is using in the power station to generate the voltage. This alternator has the effective resistance of

0.010Ω and the synchronous reactance of 0.09Ω. Determine the no-load induced voltage/phase and voltage regulation at rated current and 0.85pf lagging.

A 3-Φ, 60kVA, 415V, Υ-connected, 50Hz alternator has an armature resistance of 0.2Ω and a synchronous reactance of 2.3Ω. Determine the no-load induced voltage/phase and the voltage regulation at unity power factor.

Tests of Synchronous Generator

The Ra of the alternator is determined by the voltmeter & ammeter method. Let R be

the resistance b/w terminals. For Υ-connection, the value of the Ra is

Ra+Ra=R=0.5 →Ra=0.5R , For Δ-connection, R= Ra║(Ra+Ra) →Ra=1.5R

Open Circuit Test

To obtain the open circuit characteristics curve, the generator is driven at

no-load and its field current is set to zero. This field current is increased in

steps and the corresponding values of voltages Ef are recorded and plotted

against field current If. Figure shows the OCC curve. At a fixed value of field

current, the value of no-load voltage per phase is Ef1.

Short Circuit Test

Terminals are short with wire or ammeter to measure the SCC. The field

current is increased in steps and the corresponding armature/SCC are

recorded. The armature/SCC are plotted versus field current.

Figure shows the SCC curve. At a fixed value of field current, the value of no

load voltage per phase is Ef1 and the value of short circuit current

is Isc1.

The value of synchronous impedance is, Zs = Ef1/Isc1 and the synchronous

reactance is, Xs=√(Zs2-Ra

2).

Practice Problems

A 3-Φ, 50Hz, 33kV, Υ-connected alternator is installed. The Ra/phase is 0.54Ω. The

tests data are given: SC test: If =15A, voltage b/w lines is 415V and for OC test: If

=15A, short circuit current is 25A. Find the synchronous impedance, synchronous

reactance and full-load VR at 0.9 pf lagging.

A 3-Φ,100kVA, 50Hz, 11kV, Δ-connected alternator is installed. The per armature

resistance b/w two terminals is 0.24Ω. The tests data are : SC test: If = 25A, voltage

b/w lines is 400V and OC test: If = 25A, short circuit current is 40A. Find the

synchronous impedance, synchronous reactance & full-load VR at 0.85 pf lagging.

Power and Torque ExpressionConsider an equivalent circuit/phase to derive the relationship of power and

Torque of the alternator.

The real power & maximum power are,

P=(3VtEfsinб)/Xs,

Pmax=(3VtEf)/Xs

The torque is, T=(P/ωs)=(Pmax/ωs)sinб T= Tmax sinб

The power & torque varies sinusoid ally with a power or torque angle of б.

if we further increase the prime-mover input to the generator after a critical limit then it causes the power output to decrease.

The excess power goes into accelerating the generator, thereby increasing its speed and causing it to pull out of synchronism.

The steady-state stability limit is reached when б = 90°.

For normal steady operating conditions, the value of the power angle or torque angle is less than 90°.

Examples

The synchronous reactance of a 30MVA, 20kV, 1800rpm, 3-Φ alternator is

6Ω/phase. The induced voltage/phase is15kV and per phase terminal voltage

is 10kV. Determine the 3-Φ real power delivered to the load at power angle

of 40°, and 3-Φ maximum power.

A 30MVA, 20kV, 1800rpm, 3-Φ alternator is delivering a power of 80MW.

The induced voltage/phase is 20kV and per phase terminal voltage is 15kV.

Find the synchronous reactance if the torque angle of 40°.

Parallel Operation of Alternator

In the power station, a number of alternators are connected in parallel to generate the sufficient amount of electrical power.

The connection of the parallel alternators with the common infinite bus of National Grid of WAPDA is known as synchronization.

The following conditions should be satisfied before making parallel connection: same terminal voltage, same frequency, same phase, same phase sequence.

Figure shows 2 generator with same load/torque characteristics areconnected in parallel across the load.

Let Vt is the terminal voltage, E1 is the Induced voltage/phase by alternator1

& E2 is the induced voltage by the alternator 2. The impedances and currentsare shown in figure.

The following equations can be written as,

E1-Ia1Z1=Vt, E2-Ia2Z2=Vt, I=I1+I2,

using above relationship,

E1-Ia1Z1=(Ia1+Ia2)Z → E1=Ia1(Z+Z1)Z+Ia2Z

Similarly,

E2-Ia2Z2=(Ia+Ia2)Z → E2=Ia2(Z+Z2)Z+Ia1Z

Now the expression for the armature current using above equations.

E1(Z+Z2)-E2Z=Ia1[(Z+Z1)(Z+Z2)-Z2]

→Ia1=[(E1-E2)Z+E1Z2]/[Z1Z2+Z(Z1+Z2)]

Putting the value of in above equations, the value of and the expression of

armature current for the second alternator is,

Ia2=[(E2-E1)Z+E2Z1]/[Z1Z2+Z(Z1+Z2)]

The expression of load current is,

I=(E1Z2+E2Z1)/[Z1Z2+Z(Z1+Z2)]

An accurate approach is used for checking paralleling conditions

among the generators an apparatus called synchronoscope is used.

It is used to measure the difference in phase angle b/w the same phases of the alternators.

The alternators will be in the same phase if the needle is at zero as shown in Fig.

If the incoming alternator is faster than the running one, then the needle will move in the clockwise direction and vice versa.

Practice Problems Two alternators connected in parallel to deliver electrical power of load 5Ω.

The necessary parameters are: E1=220V, Z1=j3Ω; E2=220∟5°V,

Z2=j4Ω. Determine the Load current, Terminal voltage, power/phase

delivered by the first alternator.

Two alternators connected in parallel to deliver electrical power of load 8Ω.

The necessary parameters are: E1=210V, Z1=j5Ω; E2= 210∟5°V,

Z2=j6Ω. Find the load current, terminal voltage and 3-Φ Power delivered by

the second alternator

True, Reactive, and Apparent power in an AC Circuit

We know that reactive loads such as inductors and capacitors dissipate zero

power, yet the fact that they drop voltage and draw current gives the deceptive

impression that they actually do dissipate power. This “phantom power” is

called reactive power, and it is measured in a unit called Volt-Amps-Reactive

(VAR), rather than watts. The mathematical symbol for reactive power is

(unfortunately) the capital letter Q.

The actual amount of power being used, or dissipated, in a circuit is called true

power, and it is measured in watts (symbolized by the capital letter P, as

always).

The combination of reactive power and true power is called apparent power,

and it is the product of a circuit's voltage and current, without reference to

phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is

symbolized by the capital letter S.

As a rule, true power is a function of a circuit's dissipative elements, usually

resistances (R). Reactive power is a function of a circuit's reactance (X).

Apparent power is a function of a circuit's total impedance (Z).

True, Reactive, and Apparent power in an AC Circuit

Since we're dealing with scalar quantities for power calculation, any complexstarting quantities such as voltage, current, and impedance must be representedby their polar magnitudes, not by real or imaginary rectangular components.For instance, if I'm calculating true power from current and resistance, I mustuse the polar magnitude for current, and not merely the “real” or “imaginary”portion of the current.If I'm calculating apparent power from voltage and impedance, both of theseformerly complex quantities must be reduced to their polar magnitudes for thescalar arithmetic.There are several power equations relating the 3 types of power to resistance,

reactance, and impedance (all using scalar quantities):

True, Reactive, and Apparent Power in an AC Circuit

Please note that there are two equations each for the calculation of true and

reactive power and three equations available for the calculation of apparent

power, P=IE being useful only for that purpose. Examine the following circuits

and see how these three types of power interrelate for: a purely resistive load,

a purely reactive load and a resistive/reactive load.

True, Reactive, and Apparent Power in an AC Circuit

These three types of power -- true, reactive, and apparent -- relate to one

another in trigonometric form. We call this the power triangle:

Using the laws of trigonometry, we can solve for the length of any side

(amount of any type of power), given the lengths of the other two sides, or the

length of one side and an angle.

True, Reactive, and Apparent Power in an AC Circuit

REVIEW:

•Power dissipated by a load is referred to as true power. True power is

symbolized by the letter P and is measured in the unit of Watts (W).

•Power merely absorbed and returned in load due to its reactive properties is

referred to as reactive power. Reactive power is symbolized by the letter Q

and is measured in the unit of Volt-Amps-Reactive (VAR).

•Total power in an AC circuit, both dissipated and absorbed/returned is

referred to as apparent power. Apparent power is symbolized by the letter S

and is measured in the unit of Volt-Amps (VA).

•These three types of power are trigonometrically related to one another. In a

right triangle, P = adjacent length, Q = opposite length, and S = hypotenuse

length. The opposite angle is equal to the circuit's impedance (Z) phase

angle.