synchronous machine working and analysis

7/25/2019 synchronous machine working and analysis

1/33

Self and Mutual Inductances forSynchronous Machine with

Round Rotor

Double Layer Lap Winding

on Stator


2/33

Cross Section Diagram

a axis

baxis

caxis

qaxis

qmqa

qd

daxis

mad qqq


3/33

Stator WindingFractional Pitch

a axis

aq

q=2

rm gm

m

Pgg

2

a axis

2

mg

aq

q=4

(exaggerated end turns)

qcoils per group

2

3 mg


4/33

Self and Mutual Inductances (1)

ia(t)

ib(t)

ic(t)

a axis

daxis

qaqm

mad qqq qd


5/33

Self and Mutual Inductances (2)

cos( )

2cos( )

3

2cos( )3

aa bb cc ls A

ab bc ca s

f lf mf

af sf me

bf sf me

cf sf me

L L L L L

L L L M

L L L

L L

L L

L L

q

q

q

mme

Pqq

2

Linear Model

Balanced Winding

lsL is leakage inductance of armature phase A winding which is about10% of the maximum self inductance.

is leakage inductance of field winding.lfL


6/33

Flux Linkage (1)

( )

( )

( )

[

a aa a ab b ac c af f

aa a s b c af f

b ba a bb b bc c bf f

aa b s a c bf f

c ca a cb b cc c cf f

aa c s a b cf f

f f f af a bf b cf c

f f sf

L i L i L i L i

L i M i i L i

L i L i L i L i

L i M i i L i

L i L i L i L i

L i M i i L i

L i L i L i L i

L i L i

2 2

cos cos( ) cos( )]3 3

a me b me c mei i

q q q

At steady state, fi is DC.


7/33

Flux Linkage (2)

f

c

b

a

fmesfmesfmesf

mesfaass

mesfsaas

mesfssaa

f

c

b

a

i

i

i

i

LLLL

LLMM

LMLM

LMML

)3

2cos()

3

2cos(cos

)3

2cos(

)3

2cos(

cos

q

qq

q

q

q


8/33

Flux Linkage (3)

sAlssaas

mecmeb

measffff

mefsfcs

fcfcsaac

mefsfbs

fbfbsaab

mefsfas

fafasaaa

MLLMLL

ii

iLiL

iLiL

iLiML

iLiL

iLiML

iLiL

iLiML

)]3

2cos()

3

2cos(

cos[

)

3

2cos(

)(

)3

2cos(

)(

)cos(

)(

q

q

q

q

q

q

0 cba iiiY connected without neutral return or balanced Dconnected :


9/33

Flux Linkage (4)

f

c

b

a

fmesfmesfmesf

mesfs

mesfs

mesfs

f

c

b

a

i

i

i

i

LLLL

LL

LL

LL

)3

2cos()

3

2cos(cos

)3

2cos(00

)3

2cos(00

cos00

q

qq

q

q

q

0

cba iiiWhen


10/33

Flux Linkage in Phase A Winding

There are a total ofPgroups. These groups may be connected in series, or parallel,

or partly series and partly parallel.

cospkwcsgroupsa kqNPP Then:

csa qNPN

waa kNN

cosa a pk N

number of series turns per phase per circuit

CPP sAssume:

Note:

effective number of seriesturns per phase

per circuit on armature winding

number of parallel circuitsC

)2

cos( q apkP

BB

P

DlBpkpk

2

pk a pkN


11/33

Self Inductance of Stator Winding

If we apply current in Phase A winding, then the magnetic field for fundamentalharmonic is:

aa

a

eff

a

Pi

P

N

gB q

2cos

4 0

Now, we can calculate flux in Phase A winding from its own current.

Following the formula derived in Notes Flux Linkage in Phase Winding

pkaapkaaa NN ,,)0cos(

P

DlB pkapka

,

,

2 a

a

eff

pka iP

N

gB

4 0,

aa

eff

aa

eff

aa i

P

N

g

Dli

P

N

gP

DlN

2

00842

2

08

P

N

g

Dl

iL a

effa

aA

0

where

aa bb cc ls AL L L L L

This equation is true no matter how those P groupsof windings are connected. Note iais phase A terminal current.Na is effective number of turns connected in series per phase.


12/33

Mutual Inductance between

Stator Windings

If we apply current in Phase B winding, then the magnetic field is:

3

2

2cos

4 0 q

ab

a

eff

b

Pi

P

N

gB

Now, we can calculate flux linkage in Phase A winding from Phase B current.

)3

2cos(| ,windingBPhasefrom pkbaa N

P

DlB pkbpkb

,

,

2 b

a

eff

pkb iP

N

gB

4 0,

ba

eff

ba

eff

aa iP

N

g

Dli

P

N

gP

DlN

2

00442

2

1

2

42

0 Aa

effb

a

s

L

P

N

g

Dl

iM

where


13/33

Mutual Inductance between Stator and

Rotor Field Winding

If we apply current in rotor field winding, then when rotor is moving,the magnetic field in airgap from rotor field winding is:

Now, we can calculate flux in Phase A winding from field current.

mepkfaa N q cos

| ,windingfieldfrom

P

DlB p kfpkf

,

,

2

fme

fa

eff

f

f

eff

meaa iP

NN

g

DliP

N

gP

DlN q

q cos

842cos

2

00

)2

cos(4 0

df

f

eff

f

Pi

P

N

gB q

f

f

eff

pkf iP

N

gB

4 0,

mesfme

fa

efff

aaf L

P

NN

g

Dl

iL qq

coscos

82

0

mad qqq

where

mme

P

qq 2Define:

20

8

P

NN

g

DlL fa

eff

sf

where

meq

fwff NkN Effective number

of turns on fieldwinding.


14/33

Self Inductance of Rotor Field Winding

For the magnetic field from rotor field winding is:

Now, we can calculate flux in field winding by integrating on .

pkfff

N,

P

DlB pkfpkf

,

,

2

f

f

eff

f

f

eff

fff iP

N

g

Dli

P

N

gP

DlNk

2

00842

)2

cos(

4 0df

f

eff

fPi

P

N

gB q

f

f

eff

pkf iP

N

gB

4 0,

2

08

P

N

g

Dl

iL

f

efff

f

mf

where

dq

mflff LLL


15/33

Steady State Analysis ofRound Rotor Machine


16/33

Terminal Voltage for Round Rotor Motor

f

c

b

a

f

c

b

a

f

s

s

s

f

c

b

a

dt

d

i

i

i

i

R

R

R

R

v

v

v

v

000

000

000

000

0

cba iiiWhen

f

c

b

a

fmesfmesfmesf

mesfs

mesfs

mesfs

f

c

b

a

i

i

i

i

LLLL

LL

LL

LL

)3

2cos()

3

2cos(cos

)3

2cos(00

)3

2cos(00

cos00

q

qq

q

q

q


17/33

Round Rotor Motor at Steady State

aa s a

dv R i

dt

cosa s a sf me f L i L i q

At steady state me e r tq

2

sin

Rer

e

aa s a s sf f e e r

j

j tas a s sf f e

div R i L L I t

dt

diR i L L I e edt

Let Re ej tav e V Re ej ta Ai e I 2rj

A sf f eL I e

E

s A s A AR jX V I I E

2

08

P

NN

g

DlL

fa

eff

sf

mee

d

dt

q

s e sX L


18/33

Round Rotor Generator at Steady State

Motor Generator

s A s A AR jX V I I Es A s A AR jX V I I E

aa s a

dv R i

dt

( ) cosa s a sf me f L i L i q

2Rer

e

jj ta

a s a s sf f e

div R i L L I e e

dt

Re ej tav e V Re ej ta Ai e IAE

2

08

P

NN

g

Dl

L

fa

effsf


19/33

Open Circuit Voltage (1)

from field winding ,

,

| ( ) sin( ) cos( )a f pkA a f pk e e r a e r

d dE t N t N t

dt dt

repkfapkfaa tNN coscos| ,,windingfieldfrom

from field winding , ,| cos( ) cos( )2 2

g f pk e a r f pk a e rP PB B t B t q q

Assume the armature windings are open circuit, the magnetic field in the air gapcomes from the field winding only.

)( ret From Notes Flux Linkage in Phase WindingP

DlB pkfpkf

,

,

2

At steady state, we have 0,

dt

d pkf

)

2

cos(

)sin()(

,

,

reepkfa

reepkfaA

tN

tNtE( )

2,

rj

e a f pk N e

A

E

Phasor ofEA(t) is:


20/33

Open Circuit Voltage (2)

0

2

8 a fsf

eff

N NDlL

g P

,

,

2r pk

f pk

B Dl

P

0

4 cos( )2

fr f a me

eff

N PB Ig P q q

( )2,

rj

e a f pk N e

AE)2(

rj

fsfeA eILE

We can find out that they are the same.

( ) ( )0 2 2

,2 8 r rj ja f

A e f e f pk

eff

N NDl I e eg P

E

, 0 ,2

18 2

2

a f

A rms e f e f pk

eff

N NDlE f I

g P

, ,

f pk a f pkN


21/33

Volt-Second Balance

, , ,

,

, ,

2 4.44

or: 2 4.44

A rms e a f pk e a f pk

A rms

a f pk a f pk

e

E f N f N

EN N

f

Example: if a 60Hz generator is to be operated at 50 Hz, then the

operating voltage must be derated to 50/60 of its original value.

,f pk


22/33

Induced Phase Voltage

Therms phase voltage is, 2 4.44rms e a pk e a pk V f N f N

, cosa net a pk e net N t

For the net magnetic field, at steady state:

P

DlBpkpk

2

The steady state phase voltage in armature phase A winding is:

)2

cos()sin()( neteepkaneteepkaa tNtN

dtdtV

( )2

netj

e a pk N e

VPhasor

)2

cos( netaepknet PtBB q

pk


23/33

Y Connection (Generator)


24/33

Connection (Generator)


25/33

Example 1

4P

2 pkpk

B Dl

P

For a 3 phase, 4 pole, 24 slot, 5/6 pitch machine with double layer lap

winding, the peak magnetic filed intensity in the airgap is 0.45 T. Thereis one slot skew. The machine shaft speed is 8000 rpm. The statorinner diameter is 0.5 m. The machine length is 0.3 m. There are 10turns per coil. All the turns are connected in series. The three phasecoils are Y connected.

(1)What is the rms phase voltage of the machine?(2)What is the rms terminal voltage of the machine?

, 2 4.44rms e a pk e a pk V f N f N

a a wN N k

a cN PqN

Sq

mP

ACmachine1.m

pk


26/33

Example 2

2P

2 pkpk

B Dl

P

For this example: a cN N

(Note: This is single layer winding.)

For a simple 2 pole, 3 phase, Y connected machine (single layer

winding) shown in the figure, the peak magnetic field intensity in theairgap is 0.2T. There is no skew. The machine shaft speed is 3600 rpm.The stator inner diameter is 0.5 m. The machine length is 0.3 m. Thereare 15 turns in the coil.

(1)What is the rms phase voltage of the machine?

(2)What is the rms terminal voltage of the machine?

, 2 4.44rms e a pk e a pk V f N f N

ACmachine2.m

pk


27/33

Voltage and Speed Regulation

%100

fl

flnl

V

VVVR

Voltage regulation:

%100

fl

flnl

n

nn

SR

Speed regulation:

%100

fl

flnlSR

Or:


28/33

AC Machine Efficiency

100%out

in

P

P

out in lossP P P


29/33

AC Machine Loss Mechanism

1. Electrical or copper losses (I2Rlosses)2. Core losses

3. Mechanical losses4. Stray or miscellaneous losses


30/33

Electrical or Copper Loss

Stator Copper Loss (SCL):

Rotor Copper Loss (RCL):

23SCL A sP I R

2

RCL F FP I R


31/33

Core, Mechanical and Stray Losses


32/33

AC Generator Power Flow

power converted from mechanical to electrical

conv em mP T


33/33

AC Motor Power Flow

power converted from electrical to mechanical

conv em mP T

synchronous machine working and analysis

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