ECE 330 POWER CIRCUITS AND …...POWER CIRCUITS AND ELECTROMECHANICS LECTURE 23 SYNCHRONOUS MACHINES (3) Acknowledgment-These handouts and lecture notes given in class are based on

Copyright © 2017 Hassan Sowidan

ECE 330

POWER CIRCUITS AND ELECTROMECHANICS

LECTURE 23

SYNCHRONOUS MACHINES (3)

Acknowledgment-These handouts and lecture notes given in class are based on material from Prof. Peter

Sauer’s ECE 330 lecture notes. Some slides are taken from Ali Bazi’s presentations

Disclaimer- These handouts only provide highlights and should not be used to replace the course textbook.

11/27/2017


SALIENT POLE THREE-PHASE SYNCHRONOUS

MACHINE

• First we consider a salient pole three-phase machine as

opposed to a round rotor machine

• Salient pole machines are used in hydro-generators and

low-power single-phase synchronous motors.

• We consider a two-pole machine.

• There are three stator coils distributed so that each coil

creates a sinusoidal mmf around the periphery.

• The rotor has a field coil that carries a constant current . ri

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SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE

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The stator coils are separated mechanically. The flux current

relations can be derived as

0 2 0 2

0 2 0 2

0 2 0 2

cos 2 cos 2( 60 )

cos 2( 60 ) cos 2( 120 )=

cos 2( 60 ) cos 2( 180 )

cos cos( 120 )

a

b

c

r

L L M M

M M L L

M M M M

M M

0 2

0 2

0 2

cos 2( 60 ) cos

cos 2( 180 ) cos( 120 )

cos 2( 120 ) cos( 120 )

cos( 120 )

a

b

c

rr

iM M M

iM M M

iL L M

iM L

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• inductance matrix is symmetric.

• all inductances except are functions of

• Also and .

• All windings have the same number of turns and there is

no leakage flux.

=

a ab ac ara a

ab b bc brb b

ac bc c crc c

ar br r rr r

L M M M i

M L M M i

M M L M i

M M M L i

rL

00 =

2

LM

2 2=L M

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2 2 2 21 1 1 1=

2 2 2 2m a a b b c c r rW L i L i L i L i

a b ab a c ac b c bci i M i i M i i M

a r ar b r br c r cri i M i i M i i M

=e mWT

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2 2 2

=2 2 2

a a b b c ci dL i dL i dL

d d d

ab ac bca b a c b c

dM dM dMi i i i i i

d d d

ar br cra r b r c r

dM dM dMi i i i i i

d d d


ROUND ROTOR THREE-PHASE SYNCHRONOUS

MACHINE

For round rotor case and

2 2= = 0L M 00 =

2

LM

0 0 0

0 0 0

0 0 0

cos

cos( 120 )( ) =

cos( 120 )

cos cos( 120 ) cos( 120 ) r

L M M M

M L M ML

M M L M

M M M L

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MACHINE

2 2

0 0 0 0

2

0 0

2

1 1=

2 2

1cos

2

1cos( 120 ) cos( 120 )

2

m a a b b a c

b c c a r

b r c r r r

W L i M i i L i M i i

M i i L i i i M

i i M i i M L i

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ROUND ROTOR THREE-PHASE SYNCHRONOUS MACHINE

Since constant

In steady-state AC conditions and in terminal conditions.

Whatever happens in phase a, happens in phase b , 120 deg.

later and in phase c, 240 deg. later.

= =e m ar br cra r b r c r

W dM dM dMT i i i i i i

d d d

= sin sin( 120 ) sin( 120 )a r b r c ri i M i i M i i M

= e

m mp T

= = = = = =a b c ab ac bcL L L M M M

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MACHINE

Assume a balanced set of currents in the stator

= cosa m si I t

= cos( 120 )b m si I t

= cos( 120 )c m si I t

= =r ri I constant

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MACHINE

= [sin cos sin( 120 )cos( 120 )e

m r s sT I I M t t

sin( 120 )cos( 120 )]st

sin( ) sin( )=

2

s sm r

t tI I M

sin( 240 ) sin( )

2

s st t

sin( 240 ) sin( )

2

s st t

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MACHINE

Using the identity

We get,

, ,

sin( ) sin( 240 ) sin( 240 ) = 0s s st t t

3sin( )=

2

e m r sI I M tT

= 2m aI I =aI rms = mt

= 2 3 sin( )2

e a rm s

I IT M t t

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MACHINE

For to have an average value, , which is called

the synchronous speed.

The synchronous speed is equal to the electrical

frequency in radians per second,

where synchronous speed in rpm.

For two pole machine rpm.

eT =m s

3= sin

2

e

a rT I I M

m

s2

= = 260

sm

Nf

=sN

= 60 = 3600sN f

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VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE)

Compute va in steady state

Adding and subtracting

11/27/2017 14

00

0

= =

sin sin( 120 )2

sin( 120 ) sin( )2

a a b c ara a ab ac r

m s s m s s

m s s r s s

d di di di dMv L M M I

dt dt dt dt dt

LL I t I t

LI t I M t

m st t

0 sin2

m s s

LI t



0

3= 2 sin 2 sin( )

2 2

ra a s s s s

MIv L I t t

/20

3= [ 2 ] = 2

2

j t j tjs saa sav Re V e Re L I e e

222

jj tjr ss

M IRe e e e

11/27/2017 15

0 03= sin [sin sin( 120 )

2 2

sin( 120 )] sin( )

a m s s m s s s

s r s s

L Lv I t I t t

t I M t



0

3=

2 2

jraa s s

MIV j L I j e

0

3( )

2s sL x synchronous reactance

=22 2

j

s r s rM I e M Ij

= a ara sV jx I E

( ( ) )ar rE voltage phasor proportional to field rotor currentI

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20 ,j

a aI I e j


VOLTAGE IN STEADY-STATE (ROUND ROTOR CASE)

The equivalent circuit

Reference phasor.

Real power into the source

= a ara sV jx I E

= cos = 2 cosa m s a si I t I t

= 0a aI I

*

= [ ]a r aaP Re E I

=22

s ra a

MIP Re I

= sin2

s ra

MII

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Power in three phases:

The mechanical power output: .

Since we have DC field excitation, .

from the frequency condition , we have .

For a conservative system,

= 3T aP P

3= sin

2T s r aP MI I

= e

m mP T

= 0r

=m s r =m s

=T mP P

3= sin

2

e

a rT MI I

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POWER IN TERMS OF VOLTAGES

,

.

*

3 = 3 3P F

aIN a aaS V I V I

3

0= 3 0 ( )

a a rIN a

S

V ES V

jX

2

3

3 903 90=

a a raIN

S S

V EVS

X X

3

3 3= cos(90 ) sin

a a r a a r

IN

S S

V E V EP

X X

11/27/2017 19



Motor

Generator

Overexcitation

Underexcitation

Note: is determind by (excitation)

,

33

= sina a rINe

s s S

V EPT

X

< 0

> 02

3

33= cos

a a raIN

S S

V EVQ

X X

3 < 0INQ E cos >a r aV

3 > 0INQ E cos <ar aV

3INQ Ea r

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CONVENIENT PHASOR NOTATIONS

,

11/27/2017 21

( ) =a g ai i

= a ara sV jx I E

aar s aE jx I V

3

3= sin

a a r

T

S

V EP

X

2

3

3 3= cos

a a r aT

S S

V E VQ

X X

ar aa

s

E VI

jx



Motor

Generator

Overexcitation

Underexcitation

Note: is determined by (excitation)

,

< 0

> 0

3 < 0INQ

E cos >a r aV3 > 0INQ

E cos <ar aV

3INQ Ea r

11/27/2017 22


EXAMPLE 6.1

A three-phase wye-connected 60 Hz synchronous machine

with two poles has synchronous reactance /phase.

Operating as a motor: 30A, 254 V, PF= 0.8 leading,

windage, friction, and core losses = 400 W

Find: and , useful shaft torque, efficiency

,

= 5.0sx

arE eT

11/27/2017 23


EXAMPLE 6.1

The equivalent circuit

,

3INQ

Ear

= 254 0aV

cos = 0.8 = 36.87

= 30 36.87aI A

= 254 0 ( 5)30 36.87

= 364.3 19.23

ar a s a

ar

E V jx I

E j

V

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EXAMPLE 6.1

The torque angle is and

The torque of electric origin is

Overall efficiency

Useful shaft torque = N-m

,

3(364.3)(254)sin( 19.23 )= = 18286

5TP W

18286= = = 48.5

377

e T

m

PT N m

18286 400= =1788618286 = 97.8%

18286

17886= 47.44

377

= 19.23 = = 377 /m s rad sec

11/27/2017 25


EXAMPLE 6.4

A two-pole, three-phase, 60 Hz, wye-connected synchronous

machine, , is operating as a generator

1905 V per phase, 350 A , PF of the load is 0.8 lagging.

Find , , and the torque of electric origin

,

= 2sx

a rE

cos = 0.8 = 36.87

=1905 0aV V

= 350 36.87aI A

11/27/2017 26


EXAMPLE 6.4

,

3 sin 3(2391)(1905)(.23416)= = = 1600,000 = 1.6

2

a r a

T

s

E VP W MW

x

= =1600,000e

m mP T W

=

= 377 /

m s two pole machine

rads sec

1600,000= = 42440

377

eT N m

= = 1905 0 2(350 36.87 )

= 2325 560 = 2391 13.54

ar aa sE V jx I j

j V

13.54

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MULTI-POLE MACHINES

The number of poles in a machine is defined by the

configuration of the magnetic field pattern.

Source: machineryequipmentonline.com Source: electricaleasy.com

,

11/27/2017 28


MULTI-POLE MACHINES

The number of poles in a machine is defined by the

configuration of the magnetic field pattern.

When the instantaneous current is

in the direction indicated, the resulting

flux lines effectively create

an electromagnetic field with

North (N) and the South (S) poles

,

11/27/2017 29


MULTI-POLE MACHINE

For four slots carrying coils connected in series with

polarities indicated by dots and crosses. It can be one phase

of a three-phase winding. In this, we effectively have a four-

pole machine

Source: electronics-tutorial.ws

,

11/27/2017 30


MULTI-POLE MACHINES

For four-pole, three-phase machines the rotating field

completes two cycles (720 degrees. Electrical) in one

mechanical revolution of 360 degrees.

,

,

e m= 2lec ech

e m= = 2lec ech

dSince

dt

me =

2

echlec

p

=2

ms

p = 2 120s f

11/27/2017


MULTI-POLE MACHINES

,

In mathematical terms, the relationship between the number

of poles and synchronous speed is reflected in mutual

inductance. becomes and the frequency

condition becomes

,

,

2=

60

sm

N =

2

ms

p

=120

spNf

cosM cos2

pM

= ( ) /m s r p

= 0r=

2

ms

p

11/27/2017 32


SUMMARY FOR P POLE MACHINE

Generator action

With the generator convention for current,

In the phasor diagram

,

11/27/2017 33

= > 0T mP P

3 sin= =ar a

T m

s

E VP P

x

=ar a asE jX I V

> 0



Motor action

With the motor convention

In the phasor diagram

,

11/27/2017 34

= > 0T mP P

3 sin= =ar a

T m

s

E VP P

x

= a ara sV jx I E

< 0



Both for motor and generator

is the supply frequency

is the synchronous speed in mechanical radians per

second. the synchronous speed in rpm

,

11/27/2017 35

=2

s m

p

s

m

120s

fN

p

Documents

ECE 330 POWER CIRCUITS AND …...POWER CIRCUITS AND ELECTROMECHANICS LECTURE 23 SYNCHRONOUS MACHINES (3) Acknowledgment-These handouts and lecture notes given in class are based on