LECTURE 19 SYNCHRONOUS MACHINES (3) · 2018-04-11 · LECTURE 19 SYNCHRONOUS MACHINES (3)...
Transcript of 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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 2
Copyright © 2017 Hassan Sowidan
SALIENT POLE THREE-PHASE SYNCHRONOUS MACHINE
4/11/2018 3
Copyright © 2017 Hassan Sowidan
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
4/11/2018 4
Copyright © 2017 Hassan Sowidan
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
4/11/2018 5
Copyright © 2017 Hassan Sowidan
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
4/11/2018 6
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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 7
Copyright © 2017 Hassan Sowidan
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
4/11/2018 8
Copyright © 2017 Hassan Sowidan
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
4/11/2018 9
Copyright © 2017 Hassan Sowidan
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
4/11/2018 10
Copyright © 2017 Hassan Sowidan
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
4/11/2018 11
Copyright © 2017 Hassan Sowidan
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
4/11/2018 12
Copyright © 2017 Hassan Sowidan
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
4/11/2018 13
Copyright © 2017 Hassan Sowidan
VOLTAGE IN STEADY-STATE(ROUND ROTOR CASE)
Compute va in steady state
Adding and subtracting
4/11/2018 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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 16
20 ,j
a aI I e j
Copyright © 2017 Hassan Sowidan
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
4/11/2018 17
Copyright © 2017 Hassan Sowidan
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
4/11/2018 18
Copyright © 2017 Hassan Sowidan
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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 20
Copyright © 2017 Hassan Sowidan
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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 22
Copyright © 2017 Hassan Sowidan
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
4/11/2018 23
Copyright © 2017 Hassan Sowidan
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
4/11/2018 24
Copyright © 2017 Hassan Sowidan
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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 26
Copyright © 2017 Hassan Sowidan
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
4/11/2018 27
Copyright © 2017 Hassan Sowidan
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
,
4/11/2018 28
Copyright © 2017 Hassan Sowidan
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
,
4/11/2018 29
Copyright © 2017 Hassan Sowidan
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
Copyright © 2017 Hassan Sowidan
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
Copyright © 2017 Hassan Sowidan
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
4/11/2018 32
Copyright © 2017 Hassan Sowidan
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
Copyright © 2017 Hassan Sowidan
SUMMARY FOR P POLE MACHINE
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
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
,
4/11/2018 35
=2
s m
p
s
m
120s
fN
p