ECE 330 POWER CIRCUITS AND …...POWER CIRCUITS AND ELECTROMECHANICS LECTURE 23 SYNCHRONOUS MACHINES...

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

11/27/2017 3

The stator coils are separated mechanically. The flux current

relations can be derived as

0 2 0 2

cos 2 cos 2( 60 )

cos 2( 60 ) cos 2( 120 )=

cos 2( 60 ) cos 2( 180 )

cos cos( 120 )

L L M M

M M L L

M M M M

cos 2( 60 ) cos

cos 2( 180 ) cos( 120 )

cos 2( 120 ) cos( 120 )

cos( 120 )

iM M M

iL L M

11/27/2017 4

• 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

2 2=L M

11/27/2017 5

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

11/27/2017 6

=2 2 2

a a b b c ci dL i dL i dL

ab ac bca b a c b c

dM dM dMi i i i i i

ar br cra r b r c r

dM dM dMi i i i i i

ROUND ROTOR THREE-PHASE SYNCHRONOUS

MACHINE

For round rotor case and

2 2= = 0L M 00 =

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

11/27/2017 7

MACHINE

0 0 0 0

1cos( 120 ) cos( 120 )

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

11/27/2017 8

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

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

m mp T

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

11/27/2017 9

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

11/27/2017 10

MACHINE

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

m r s sT I I M t t

sin( 120 )cos( 120 )]st

sin( ) sin( )=

s sm r

t tI I M

sin( 240 ) sin( )

s st t

sin( 240 ) sin( )

s st t

11/27/2017 11

MACHINE

Using the identity

We get,

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

3sin( )=

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

11/27/2017 12

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

a rT I I M

= = 260

= 60 = 3600sN f

11/27/2017 13

VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE)

Compute va in steady state

Adding and subtracting

11/27/2017 14

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

3= 2 sin 2 sin( )

ra a s s s s

MIv L I t t

3= [ 2 ] = 2

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

jj tjr ss

M IRe e e e

11/27/2017 15

0 03= sin [sin sin( 120 )

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

jraa s s

MIV j L I j e

2s sL x synchronous reactance

s r s rM I e M Ij

= a ara sV jx I E

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

11/27/2017 16

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

s ra a

MIP Re I

= sin2

11/27/2017 17

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

m mP T

=m s r =m s

=T mP P

3= sin

a rT MI I

11/27/2017 18

POWER IN TERMS OF VOLTAGES

3 = 3 3P F

aIN a aaS V I V I

0= 3 0 ( )

a a rIN a

V ES V

3 903 90=

a a raIN

3 3= cos(90 ) sin

a a r a a r

V E V EP

11/27/2017 19

Generator

Overexcitation

Underexcitation

Note: is determind by (excitation)

= sina a rINe

33= cos

a a raIN

3 < 0INQ E cos >a r aV

3 > 0INQ E cos <ar aV

3INQ Ea r

11/27/2017 20

CONVENIENT PHASOR NOTATIONS

11/27/2017 21

( ) =a g ai i

= a ara sV jx I E

aar s aE jx I V

3= sin

3 3= cos

a a r aT

V E VQ

Generator

Overexcitation

Underexcitation

Note: is determined by (excitation)

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

= 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

E V jx I

11/27/2017 24

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

18286= = = 48.5

PT N m

18286 400= =1788618286 = 97.8%

17886= 47.44

= 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

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

E VP W MW

= =1600,000e

m mP T W

= 377 /

m s two pole machine

rads sec

1600,000= = 42440

eT N m

= = 1905 0 2(350 36.87 )

= 2325 560 = 2391 13.54

ar aa sE V jx I j

11/27/2017 27

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

echlec

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

cosM cos2

= ( ) /m s r 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

E VP P

=ar a asE jX I V

Motor action

With the motor convention

In the phasor diagram

11/27/2017 34

= > 0T mP P

3 sin= =ar a

E VP P

= a ara sV jx I E

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

ECE 330 POWER CIRCUITS AND …...POWER CIRCUITS AND ELECTROMECHANICS LECTURE 23 SYNCHRONOUS MACHINES...

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