Chapter 5

Synchronous Generator

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Home Work Assignment

Problems: 5.1, 5.2 (exclude part (e), 5.4, 5.5, 5.7, Due Next week on Monday.

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Synchronous Gen. Introduction! In Synchronous generator, DC current is

supplied to the rotor winding, which produces rotor magnetic field.

! The rotor is then rotated by prime mover(diesel engine, steam turbine or water turbine), producing a rotating magnetic field in the machine.

! The rotating magnetic field induces a three phase set of voltages with in the stator windings of the generator.

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Synchronous Generator Rotor Construction

Rotor is a large electro magnet with a DC winding on it. The poles are of two types:

1. Salient or Protruding or Sticking out of the surface of the rotor.

2. Non Salient or flushing with the rotor surface.

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! Thin laminations same like transformer to reduce eddy currents.

! DC current must be supplied to the rotor

Salient Pole Rotor4 or more poles

Non salient Pole RotorNormally used for 2 or 4 poles

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! How to supply DC to the rotor/field1. By an external DC source by means of slip

rings and brushes. 2. From a special DC source, mounted directly

on the shaft." Slip rings are used for smaller machines

Disadvantages of Slip rings.1. Slip rings increase maintenance, need to

check brushes regularly. 2. Carbon brushes has voltage drop,

significant power loss.

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" For larger machines Brushless exciters are used.

" It is a small AC generator with its field circuit on the stator and armature circuit on the rotor.

" Three phase output generated on the rotor is then rectified and fed to the rotor field circuit.

Brushless Exciters

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5.2. Speed of rotation of synchronous machine

! Electrical frequency is locked in or synchronized with the mechanical rate of the rotation of generator.

! To generate60 Hz in 2 pole nm = 360050 Hz in 2 poles nm = 3000

120Pnf m

e =

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5.3: Internal Generated Voltage

Internal generated voltage is given by:

Where

IF ω in electrical rad/secIF ω in mechanical rad/sec

φω

φ

φ

KE

fNfNE

A

c

cA

=

∏=

∏=

22

2

2

2

2

PNK

NK

f

c

c

=

=

∏=ω

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Internal Generated Voltage

φωKE A =Where K is machine design constant, Ø is flux and ω is frequency

speedfluxEA &∝⇒

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If and Ø are related by magnetization curve:

Ø

IF

fA

fA

IEIE

&& &

∴

∝ φφ

can be related for a fixed ω

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Magnetization Curve for EA and If

EA

IF

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5.4. The Equivalent Circuit of Synchronous Generator

EA � Internal voltage generatedVØ � Output voltage at generator terminals.EA = VØ, Only when there is no armature current

i.e., no load condition.In reality EA ≠ VØ because:

1. self inductance of armature coil2. Resistance of armature coil3. Distortion of air gap magnetic field by current

flowing in stator.4. The effect of salient pole rotor shape.( no

discussion in this course)

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! When generator spins, EA is induced in stator windings.

! If loaded current flows in the stator which will produce a magnetic field.

! This stator magnetic field distorts the original rotor magnetic field. The effect is called Armature Reaction which can be modeled by X:

! Estat lags IA by 90°! Self inductance of the stator coil can be modeled

by LA => reactance XA

Or Total reactance is given by XS=XA+X

Astat jXIE −=

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! Resistance of winding RA

=>

We can draw the full equivalent circuit.

AAASA IRIjXEV −−=φ

5.4: The Final Equivalent Circuit of Synchronous Generator

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17

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Since the three phases of synchronous generators are identical in all respects except for phase angle. This leads to use of per phase equivalent circuit.

IA

+

-DC

jXs RA

EA1

When it is true?- Only if the three phases loads are balanced.

Per Phase Equivalent Circuit

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Phasor Diagram

IA Vφ IARA

jXSIA

Unity power factor

Vφ

EA

IARAIA

jXSIA

IA

Vφ

EAjXSIA

IARA

Lagging Power Factor

Leading Power Factor

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For a given : EA Required for lagging p.f. > EA required for leading p.f.

Equivalently (or we can say that), For given field current and magnitude of

load current :Terminal voltage for lagging load < Terminal

voltage for leading load.

In real machine, therefore some times RA can be neglected.

φV

ARX >>

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Mechanical power Electrical Energy(Prime Move)The speed is almost constant regardless of power demand. If the speed is not constant the system�s frequency would wander.

→

Power and Torque in Synch. Machine

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Pin=Γappωm Pconv= Γindωm Pout=√3VTωI2

cosθ

Stray losses Friction and windagelosses

Core losses

I R losses2

Power Flow Diagram Synch. Generator

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mindconvP ωτ=

γCosIEP AAconv 3=Angle between EA and IAγ

θθ

φ CosIVPCosIVP

Aout

LTout

33

==

θθ

φ SinIVQoutSinIVQout

A

LT

33

==

XS >> RA therefore can neglect RA

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rδδδδ

θθθθ

r

Vφφφφ

jXSIA

EA

θθθθ

θθθθ

EASinδδδδ=XSIACosθθθθ

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P depends upon when = 90δ δ

S

A

XEV

P ϕ3max = (Static Stability Limit of the

generator)

S

A

S

AA

S

AA

AAS

XSinEV

P

XSinEV

CosIV

XSinECosI

SinECosIX

δ

δθ

δθ

δθ

ϕ

ϕϕ

3

33

=

=

=

=

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Normally full load torque angle is 15° to 20°

S

AA X

SinEVCosIVP

δθ ϕ

ϕ

33 ==

For fixed φV

θδθ

SinIQSinEP

CosIP

A

A

A

∝∝∝

&

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Sm

Aind

mindconv

XSinEV

P

ωδ

τ

ωτ

ϕ3=⇒

=

Induced torque in terms of electrical quantities

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5.7: Measuring Synchronous Generator Model Parameters

Need to Find! Relationship between field current &

flux (or between If and EA because EA α φ).

! The Synchronous reactance.! The armature resistance.

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1. Run the generator at rated speed. 2. Terminals are disconnected from the load.3. Set the field current to zero.4. Increase the field current gradually in

steps and measure the terminal voltage.5. Since IA = 0, terminals are open therefore

EA = Vφ6. Construct a plot of EA or VT versus If from

this function.- The plot is called Open Circuit Characteristic.

Open Circuit Test

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- Given If you can find the VT generated.

Air gap line

(OCC)

VT (v)

If (A)

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1. Adjust the field current to zero, short circuit the terminals of the generator through an ammeter.

2. Increase If and measure IA (armature current or line current).

Short Circuit Test

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SA

AA jXR

EI+

=

22SA

AA

XR

EI+

=

Vφφφφ=0v

jXsIA

EA

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When Vφ = 0Machine internal impedance =

Since XS >> RA

If IA and EA are known XS can be calculated1. EA from O.C.C at a given If2. IA,sc at that If3. Find XS (Approx.) by:

22SAS XRZ +=

A

A

IE

=

AA

AS I

ocVIEX ϕ=≈

A

AS I

EX =

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" The winding resistance RA can be approximately calculated by applying DC voltage to the winding while machine is stationary and measure the resulting current.

" Since DC voltage therefore the reactance of winding will be zero.

Short Circuit Ratio" The ratio of field current required

for rated voltage at open circuit to the field current required for rated armature current for short circuit.

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OCC

SCC

Xs

Air gap line

If

VT, V

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5.8: Effect of load changes on synchronous generator operating alone

-In this analysis we will ignore the resistance RA.

Case I Given that

Load

Generator

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! Field current is Constant and prime mover keeps the speed constant.

is constant

What is effect of load changes?1. Operating at lagging p.f and |IA|

increases but angle is same (same p.f)

(RA neglected)If lagging loads are added (+Q or inductive

reactive power loads). Vφ decreases significantly.

AE⇒ ωϕ ..KE A =

ASA IjXVE += ϕ

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EA’

EA

jIAXS

Vφ

δ’ δ

θ

IA

ResultWhen load increases Vφ decreases.

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EA’EA

δ'δ

IA

I’A

δ'δ

jIAXS

V'φVφ

IA

2.Generator Loaded with unity p.f. and IAincreases:

If unity p.f, no reactive powersare added to generator,there is a slight decreasein Vφ and the

terminal voltage.

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3. Generator Loaded with leading p.f:! If leading loads (� Q or capacitive

reactive power loads) are added to generator, Vφ and the terminal voltage will rise.

! A convenient way to compare the voltage behavior of two generators by their voltage regulation

100×−

=nl

flnl

VVV

VR

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! Synchronous generator operating at1. Lagging p.f. has fairly large

positive voltage regulation.2. Unity p.f. has small positive voltage

regulation.3. Leading p.f. has negative voltage

regulation.

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! If load varies, change EA to keep Vφconstant

ωϕ ..KE A =

How to keep Vφ Constant when load changes

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Change K or ω? (No, why?)# K machine design# Frequency should not change in normal system,

So change φf by If.

E’A

jI’AXS

Vφ

jIAXS

EA

IA

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Case Study

! Lagging p.f. and IA increases.! This will decrease Vφ! To bring Vφ back to its original value

1. Decrease Rf, this will increase If.2. Increase in If will increase φf

3. Increase in φf increases 4. When EA increases Vφ increases.

ωϕ ..KE A =

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ϕωKEA =

- This process can be reversed when the load decreases.

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1. Several generators can supply bigger loads than one machine itself.

2. Increase reliability (what is reliability?). - If one fails the total power is not lost.

3. If many generators - can shut down one and do the maintenance.

Advantages Of Parallel Operation of AC Generators

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4. If only one generator, then when operating at low load it is inefficient.-When several generator operates then at low load can use few of them running at higher load instead of running all the generators at low load thus making the over system more efficient.

Advantages Of Parallel Operation of AC Generators

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! The two generators must have1. Same rms line voltages2. Same phase sequence3. Same phase angle4. Almost same frequency.Then the two generators can be put in

parallel operation.

Paralleling Conditions