Lecture NotesEEE 360 TOPIC 4 Synchronous Machine - Lecture 13 docx

Machine• 9 SYNCHRONOUS MACHINES Round Rotor Machine • The stator is a ring shaped laminated iron-core with slots.. SYNCHRONOUS MACHINESRotor generated Flux and Induced Voltage, Round

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01/19/24 360 Topic 6 Synchr Machine

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Lecture 13

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Animation 1

Voltage induced

in a rotating loop

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Energy Conversion

Concept:

• Generators convert mechanical energy to electric energy.

• Motors convert electric energy to mechanical energy.

• The construction of motors and generators are similar

• Every generator can operate as a motor and vice versa.

• The energy or power balance is :

– Generator: Mechanical power = electric power +

losses

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– L is the length of the loop

– D is the width of the loop

– B is the magnetic flux

density

– n is the number of turns

per seconds

D B

L

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• The magnetic flux through

the loop changes by the

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• The change of flux linkage

induces a voltage in the loop

 t 

sin L

D B

N dt

t cos d

L D B

N dt

t d N t

1 B

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• The voltage is sinusoidal

• The rms value of the

induced voltage loop is:

• View the animation of voltage generation

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SYNCHRONOUS MACHINES

Round Rotor Machine

• The stator is a ring

shaped laminated

iron-core with slots.

• Three phase windings

are placed in the slots.

• Round solid iron rotor

with slots.

• A single winding is

placed in the slots Dc

current is supplied

through slip rings.

Concept (two poles)

A - B

B - A

B

C

Stator with laminated iron-core

Slots with winding

Rotor with dc winding

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Salient Rotor Machine

• The stator has a laminated

iron-core with slots and three phase

windings placed in the slots.

• The rotor has salient poles

excited by dc current.

• DC current is supplied to the rotor

through slip-rings and brushes.

• Concept (two poles)

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-01/19/24 360 Topic 6 Synchr Machine

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Construction

• The picture shows the laminated

iron core and the slots (empty

and with winding).

• The winding consists of copper

bars insulated with mica and

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SYNCHRONOUS

MACHINES

Stator details

• Coils are placed in slots

• Coil end windings are

bent to form the

armature winding.

Slots Coil

Iron core

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Round rotor

• The round rotor is used

for large high speed

(3600rpm) machines.

• A forged iron core (not

laminated,DC) is

installed on the shaft.

• Slots are milled in the

iron and insulated

copper bars are placed

in the slots

• The slots are closed by

wedges and re-enforced

with steel rings

Round rotor

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Rotor Details

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SYNCHRONOUS

MACHINES

Salient pole rotor construction

• The poles are bolted to the shaft.

• Each pole has a DC winding.

• The DC winding is connected to the slip-rings (not shown).

• A DC source supplies the winding with DC through brushes

pressed into the slip ring

• A fan is installed on the shaft to assure air circulation and

effective cooling

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SYNCHRONOUS

MACHINES

Construction

• Low speed, large

hydro-generators may have more

than one hundred poles

• These generators are

frequently mounted vertically

• The picture shows a large,

horizontally arranged

machine.

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Animation 2

Generator Operation

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Test 2 Transformer

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Lecture 14

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Operation concept

• The rotor is supplied by DC current If

that generates a DC flux f.

• The rotor is driven by a turbine with

a constant speed of ns.

• The rotating field flux induces a

voltage in the stator winding

Operation (two poles)

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SYNCHRONOUS

MACHINES

• The frequency - speed relation is f = (p / 2) ns = p ns / 2

p is the number of poles.

• Typical rotor speeds are 3600 rpm for 2-pole, 1800 rpm for 4 pole and

E  E bn E rms ei120deg E cn E rms e i240deg

w f a f

a w

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• At no load condition, the induced

voltage is equal to the terminal

voltage.

• When the generator is loaded,

the induced voltage drives a

current I a through the load

• The load current produces a flux

ar that reduces the field flux

• The armature generated flux has

constant amplitude and rotates

Operation (two poles)

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• The armature flux induces a voltage

E s in the stator winding

• This voltage is subtracted vectorially

from the field induced voltage The

terminal voltage is : V t = E f - E s .

• The E s voltage can be represented

by an equivalent armature reactance

times the armature current

E s = I a j X a

• Operation (two poles)

Armatureflux ar

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• The Es voltage can be represented by an equivalent armature

reactance times armature current, Ia.

• The reactance is:

ar N a = L ar I a L ar = ar N a / I a

X ar =  L ar =  ( ar N a / I a.)

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• The equation permits the development of an equivalent circuit, that

consists of a voltage source E f and a reactance X a r connected in series

• The stator winding has resistance and some leakage inductance that is added to armature reactance

• The X a r + X l eakage is called synchronous reactance X sy n.

• The value of the synchronous reactance is more than 100%.

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• Equivalent circuit: V t = E f - I a j X syn

Rs

jXsyI

Ef



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• Calculate the synchronous reactance in ohm

• Calculate the rated current and the line to ground terminal voltage

• Draw the equivalent circuit

• Calculate the induced voltage, Ef, at rated load and pf = 0.8 lag

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Rated current and line-to-ground terminal voltage

Terminal voltage is: V gn 24kV

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Equivalent circuit for the numerical exercise

• Results of the calculation

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Rotor generated Flux and Induced

Voltage, Round rotor machine

• For the calculation of the induced

voltage, the machine is simplified

• On the stator windings of each

phase is represented by one

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Round Rotor generated flux

• The rotor generated field is

constant along the gap The field is

calculated by the Ampere law.

• The upper part of the rotor the B

lines going out, the lower part

entering into the rotor.

Magnetic axis of the rotor

N I

H B

gH 2

dl H

N

I

f f

o f

f f

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• The flux density B distribution

along the rotor surface is a

rectangular waveform.

approximated by its Fourier series

only the base harmonics is

considered The base harmonics is:

• The B has sinusoidal space

 m 0 360

1.5 1 0.5 0 0.5 1 1.5

B m  m

B  m

 m

m f

f o m

f

g 2

N I

4 cos

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• 39

• The rotor and with the rotor the

flux rotates with an angular

speed of 

• The flux density at

• The flux links with coil A + - A - is

the integral of the flux density B

) (

cos B

) t , (

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N S

All flux links with phase A,

when  is 00 deg

The flux linkage is zero, when  is +/-900

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• The flux links with coil A+- A- is:

• Maximum flux is:

• where: r is the rotor radius, L rotor length, N f number of turns in the

rotor, I f is the field current, n s is the synchronous speed

  = 2  p n s / 2  m = t

) (

cos r 2 L B

4 d

r L ) cos(

B

4

m f

2

m f

I N 4 D L B

f max

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Rotor induced stator voltage (Round rotor machine)

• The induced voltage in phase A is:

• The rms induced voltage is: A fA max

max fA A

L g

I N N

t cos D L

B dt

d N )

t

cos(

dt

d N dt

d N

E

max fA A

f f o A

f A

max fA A

A f A fA

4

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max fA A

fA

N f 44 4

2

N E

g 2

I N B

f max

fA

f f o f

• Pole flux max base component

• Calculate Induced voltage

• Steps of induced voltage calculation:

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Student class room exercise

• A two pole, three-phase, wye connected, round rotor synchronous generator data:

• Stator (armature): Length L = 1.2 m, Number of turns per phase

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Lecture 15

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Animation 3

Rotating Flux

Generation

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Load Current generated flux

• The load current of phase A

generates an ac flux

A(t) = max sin(t)

• This flux generates a pulsating

field that changes from Max to

zero to negative Max

• The direction of flux is

perpendicular to the A phase

winding (direction by the right

hand rule).

• The length of the flux vector on

the figure, if t =  o , is:

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• The load current of phase B

generates an ac flux.

B (t) =  max sin(t -120 o )

perpendicular to the B phase

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• The load current of phase C

generates an ac flux

C (t) =  max sin(t-240 o )

perpendicular to the C phase

hand rule

• The length of the flux vector on

the figure, if t =30 0, is:

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Load current generated flux

• Total flux is the vector sum of

the three components:

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• The resultant flux

amplitude is 1.5 times the

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Load current generated flux

• The comparison of the two figure shows:

– The amplitude of the flux is the same in both figure, but the

angle has advanced by 300.

– The three phase load current produces a rotating flux.

– The amplitude of the resultant flux is constant and 3/2 times the pole flux.

– The speed of the flux is the synchronous speed

• For the demonstration of the rotating field open the “Empty

Stator Fed by 3 Phase Power Supply” animation program.

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Load current generated flux calculation :

• The A phase load currents generated magnetomotive forces (mmf) at a selected mline in the air gap is:

– mmfAA  cos mcostcos m

• Similarly the mmf of phase B and C are:

– mmfB = IB N cos (m-120) = I N cos (t - 120) cos (m-120)

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• The three phase current generated mmf is the sum of the above three

• The mmf for concentric winding is : mmfmax = I N

• The mmf for distributed winding is : mmfmax = 2 I N / 

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• The three-phase load currents generate a mmf that is described by:

mmf (t) = (3 /2) mmfmax sin (m - t )

.

• According to this equation the three phase generated mmf is the

projection of the (3/2) mmf max vector on the m

line at any time and position m.

• Also the equation describes a

rotating mmf, that produces

a rotating flux.

t

mmagnetic axis

3 / 2 IN

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• The rotor generated field is

constant along the gap The

field is calculated by the

H B

H g dl

H N

I

A A

o A

A A

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SYNCHRONOUS

MACHINES

• The flux density B distribution along the rotor surface is a rectangular

waveform

• The rectangular wave form is approximated by its Fourier series

• The higher harmonics are neglected, only the base harmonics is

considered The base harmonics is:

• This assumes that the B A0 has sinusoidal space distribution The A

m A

A o

m A

o m

A 0

g 2

N I

4 cos

H

4 cos

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SYNCHRONOUS

MACHINES

• The flux generated by phase A is the integral of the flux

density B The integral of the cos function is 2

• L is the length, r is the radius, D = 2r is the diameter of the machine

• A sinusoidal AC current supplies Phase A that results in an Ac

time varying flux:

D L B

4 r

2 L B

4 d

r L cos

B

4

A A

m 2

2

m A

) t sin(

D

L g

N I 4

) t sin(

D L B 4

A max

A

max A

A o

A A

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SYNCHRONOUS

MACHINES

• Similarly phase B and C will also generates a time varying ac flux

• These fluxes are shifted by 120o or 240o respectively

• The sum of the three flux results in a rotating flux with an amplitude 3/2

times the phase generated flux

• The amplitude of the rotating flux is :

D

L g

N I

4 2

3 LD B

4 2

3 2

o A

max A max

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SYNCHRONOUS

MACHINES

• This flux is represented by an equivalent armature inductance using

the following equation:

• The armature winding has leakage inductance The sum of the leakage inductance and armature inductance gives the synchronous inductance However the leakage is negligible in most cases:

I

N L

and I

L

N phase max  armature armature  phase max

D

L g

N

4 2

3 I

N X

L I

N )

L L

( L

X

A o

max phase A

synch

leakage

max phase A

leakage armature

synch synch

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A max

phase

A max

fA

A A o A

N X

D L B

4 2 3

D L B 4

g 2

I N B

Calculate

• Pole generated B

• Base component of A phase

generated flux

• Three phase generated flux

Steps of synchronous reactance calculation:

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Student class room exercise

• A two pole, three phase, wye connected, round rotor synchronous generator data are:

• Stator (armature): Length L = 1.2 m, Number of turns per phase

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Lecture 16

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Power angle Characteristics Round Rotor Machine

• A synchronous machine supplies an electric network with constant voltage under steady state conditions

• The terminal voltage in the machine is kept constant by the regulation

of the field current

• The generator speed is constant, at the synchronous speed determined

by the network frequency and the number of poles in the machine.

• An increase of input mechanical power increases the torque Calculate the output power variation with the input power

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NetworkBus



Network

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• The complex power delivered by the generator is:

• After simplification we get:

Generator

NetworkBus

tn fn s

tn

fn

X

V cos

X

V E j

sin X

V E

S

2

33

s s

tn

i fn tn

X i

V e

E V

I V

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SYNCHRONOUS

MACHINES

• The real and reactive power are

• The real power is maximum if = 900

• The maximum torque is:

tn fn s

tn fn

X

V cos

X

V E j

Q

sin X

V E

P

2

33

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Power angle Characteristics

• The P() curve shows that the

increase of power increases the

angle between the induced voltage

and the terminal voltage

• The power is maximum when  =90 o

• The further increase of input power

forces the generator out of

synchronism This generates large

current and mechanical forces.

• This angle corresponds to the angle

between the field flux and the stator

generated rotating flux

Round Rotor Machine

0 20 40 60 80 100

P(t) = ) 



Pmax

Tiêu đề	Synchronous Machine
Tác giả	George G. Karady
Trường học	Not Specified
Chuyên ngành	EEE 360
Thể loại	Lecture Notes
Năm xuất bản	2024
Thành phố	Not Specified

Định dạng
Số trang	77
Dung lượng	4,08 MB

Lecture Notes EEE 360 TOPIC 4 Synchronous Machine - Lecture 13 docx

Lecture NotesEEE 360 TOPIC 4 Synchronous Machine - Lecture 13 docx