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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
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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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SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE
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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SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE
• 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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SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE
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
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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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ROUND ROTOR THREE-PHASE SYNCHRONOUS
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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ROUND ROTOR THREE-PHASE SYNCHRONOUS
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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ROUND ROTOR THREE-PHASE SYNCHRONOUS
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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ROUND ROTOR THREE-PHASE SYNCHRONOUS
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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ROUND ROTOR THREE-PHASE SYNCHRONOUS
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
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VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE)
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
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VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE)
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
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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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VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE)
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
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POWER IN TERMS OF VOLTAGES
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
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POWER IN TERMS OF VOLTAGES
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
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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
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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
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
,
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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
,
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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
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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
,
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= > 0T mP P
3 sin= =ar a
T m
s
E VP P
x
=ar a asE jX I V
> 0
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Copyright © 2017 Hassan Sowidan
SUMMARY FOR P POLE MACHINE
Motor action
With the motor convention
In the phasor diagram
,
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= > 0T mP P
3 sin= =ar a
T m
s
E VP P
x
= a ara sV jx I E
< 0
![Page 35: 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](https://reader036.vdocuments.net/reader036/viewer/2022062403/5fddc3aa6c063811123ae13c/html5/thumbnails/35.jpg)
Copyright © 2017 Hassan Sowidan
SUMMARY FOR P POLE MACHINE
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