lecture 19 synchronous machines (3) · 2018-04-11 · lecture 19 synchronous machines (3)...

Copyright © 2017 Hassan Sowidan

ECE 330

POWER CIRCUITS AND ELECTROMECHANICS

LECTURE 19

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.

4/11/2018


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

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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

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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

4/11/2018 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

,

4/11/2018 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

4/11/2018 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

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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

4/11/2018 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

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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

,

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MULTI-POLE MACHINES

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

,

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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

,

4/11/2018 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

4/11/2018


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

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SUMMARY FOR P POLE MACHINE

Generator action

With the generator convention for current,

In the phasor diagram

,

4/11/2018 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

,

4/11/2018 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

,

4/11/2018 35

=2

s m

p

s

m

120s

fN

p

lecture 19 synchronous machines (3) · 2018-04-11 · lecture 19 synchronous machines (3)...

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