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SYNCHRONOUS MACHINES
Two-axis model
du
f u
f i
di
Di
Qi
quqi
γ
d
q
d
q
Q D
f
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Synchronous machines
8 MW diesel-generator 270 MVA turbo-generator
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Rotor windings
Field winding Damping bar Field winding Damping bar
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PM synchronous machine; 1 MW ship propulsion motor
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ψ ωψ
ψ ωψ
ψ
ψ
ψ
⎧= + −⎪
⎪⎪ = + +⎪⎪⎪
= +⎨⎪⎪ = +⎪⎪⎪ = +⎪⎩
d
d
ddd
d
d0d
d0
d
dd s d q
qq s q d
f f f f
DD D
QQ Q
u R it
u R it
u R it
R it
R it
ψ
ψ
ψ
ψ
ψ
⎧ = + +⎪
⎪ = + +⎪⎪
= + +⎨⎪
= +⎪⎪
= +⎪⎩
d d d df f dD D
f df d f f fD D
D dD d fD f D D
q q q qQ Q
Q qQ q Q Q
L i L i L i
L i L i L i
L i L i L i
L i L i
L i L i
Voltage, flux-linkage and motion equations
( ) ω
ψ ψ
γ ω
⎧
− = +⎪⎪⎨⎪ =⎪⎩
3 d
2 d
d
d
q q md d
J p i i T p t
t
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Space-vector model for induction motor
Rotor frame of reference
Voltage and flux linkage
The axes in the rotor frame of reference denoted by d and q
ψ ωψ
ψ
⎧= + +⎪
⎪⎨⎪
= +⎪⎩
d j
dd
d
r sr r r
s s s sr r r r
r r r
u R it
u R it
ψ
ψ
⎧ = +⎪⎨= +⎪⎩
r r r s s m r s
r r r m s r r r
L i L i
L i L i
⎧ = +⎪⎨
= +⎪⎩
r s sd sq
r r rd rq
i i i
i i i
ψ ωψ
ψ ωψ
ψ
ψ
⎧= + −⎪
⎪⎪ = + +⎪⎪⎨⎪ = +⎪⎪⎪ = +⎪⎩
d
dd
dd
dd
d
sdsd s sd sq
sq
sq s sq sd
rdrd r rd
rq
rq r rq
u R it
u R i t
u R it
u R i t
ψ
ψ
ψ ψ
= +⎧⎪ = +⎪⎨
= +⎪⎪ = +⎩
sd s sd m rd
sq s sq m rq
rd m sd r rd
rq m sq r rq
L i L i
L i L i
L i L iL i L i
i s d
i s d
u s d
u s d
i s q
i s q
u s q
u s q q
d
β
α
( )ψ ψ = −
3
2e sd sq sq sdT p i i
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Load angle
γ
ω =
d
dt
δ
ω ω − =
d
ds tis replaced by
where the load angle δ is the angle
between the excitation voltage vector u p (on q-axis) and stator voltage
vector ur s
δ
δ
=⎧⎪⎨ =⎪⎩
ˆ sin
ˆ cos
d s
q s
u u
u u
For synchronous machines, the position angle of the rotor is defined
differently from the angle γ typically used for induction machine analysis
γ
d
q
r su
pu
δ
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Direct-axis equivalent circuit
= = =df dD fD mdL L L L Assumption
where Lmd is the direct-axis magnetising inductance
ψ ωψ
ψ
ψ
⎧= + −⎪
⎪⎪
= +⎨⎪⎪
= +⎪⎩
d
dd
d
d0 d
dd s d q
f f f f
DD D
u R i t
u R it
R i t
( )
( )
( )
σ
σ
σ
ψ
ψ
ψ
⎧ = + + +⎪⎪
= + + +⎨⎪
⎪ = + + +⎩
d d d m d f D
f f f m d f D
D D D m d f D
L i L i i i
L i L i i i
L i L i i i
( )
( )
( )
σ
σ
σ
ωψ ⎧
= − + + + +⎪
⎪⎪
= + + + +⎨⎪⎪
= + + + +⎪⎩
d d
d dd d
d dd d
0
d d
dd s d q d m d f D
f f f f f m d f D
DD D D m d f D
iu R i L L i i i
t tiu R i L L i i i
t ti
R i L L i i i
t t
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Direct-axis equivalent circuit
di
ψ d
dd
td
dmdL
t
σ
d
d
DL
t
σ
d
d
f L
t
DR f R
+ +d D f i i i Di f i
σ dd
dLt
f u
( )
( )
( )
σ
σ
σ
ωψ ⎧ = − + + + +⎪⎪⎪
= + + + +⎨⎪⎪
= + + + +⎪⎩
d dd d
d d
d d
d d0 d d
ds q m Dd d d d f
f m D f f f f d f
DD D D m Dd f
iu R i L L i i it t
iu R i L L i i i
t t
iR i L L i i it t
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More accurate equivalent circuit for direct axis
Assumption
where Lmd is the direct-axis magnetising inductance
di
ψ d
dd
td
dmdL
t
σ dd
DLt
σ dd
f Lt
DR R
+ +d D f i i i Di f i
σ
d
ddL
t f uσ d
dkL
t
= =
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Quadrature-axis equivalent circuit
Notation
where Lmq is the quadrature-axis magnetising inductance
=qQ mqL L
ψ ωψ
ψ
⎧= + +⎪⎪
⎨⎪ = +⎪⎩
d
dd
0
d
qq s q d
QQ Q
u R it
R i
t
( )
( )
σ
σ
ψ
ψ
⎧ = + +⎪⎨⎪ = + +⎩
q q q m q Q
Q Q Q m q Q
L i L i i
L i L i i
( )
( )
σ
σ
ωψ ⎧
= + + + +⎪
⎪⎨⎪ = + + +⎪⎩
d d
d dd d
0d d
qq s q d q m q Q
QQ Q Q m q Q
iu R i L L i i
t ti
R i L L i it t
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Quadrature-axis equivalent circuit
qi
ψ d
d
q
td
dmqL
t
σ dd
QLt
QR
+q Qi i Qi
σ
d
dqL
t
( )
( )
σ
σ
ωψ ⎧
= + + + +⎪⎪⎨
⎪ = + + +⎪⎩
d d
d dd d
0 d d
qq s q d q m q Q
QQ Q Q m q Q
iu R i L L i i
t ti
R i L L i it t
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Notations and abbreviations
Leakage factors and other abbreviations
σ
σ
σ
σ
= −
= −
= −
= −
2
2
2
2
1
1
1
1
df df
d f
dDdD
d D
fD fD
f D
qQqQ
q Q
L
L L
L
L LL
L L
LL L
µ
µ
= −
= −
1
1
df fD f
dD f
dD fDD
Ddf
L L
L L
L L
L L
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Synchronous machine in steady state
Steady state => a) Space vectors are constants in rotor frame
of reference
= + =
= + =
j constant
j constant
r s d q
r s d q
u u u
i i i
b) The time-derivatives of flux linkages vanish
ψ ψ = =
dd0
d d
qd
t t
c) The currents of the damper windings are zero
= = 0D Qi i
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Space-vector diagram for
a synchronous generator
r si
r su
pu
ω df f L i
ω − q qL i
ω j d dL i
ωψ j r s r s sR i
f id
q
ψ r s
ϕ
δ
di
qi
( )ω − -j r q sdL L i
ω -j r q sL i
ωψ
ωψ
⎧ = −⎪
= +⎨
⎪ =⎩
s qd d
q s q d
f f f
u R i
u R i
u R i
ψ
ψ
= +⎧⎪⎨
=⎪⎩
d d d df f
q q q
L i L i
L i
ω ω
ω
= + − +
= =
j
ˆ j j
r r s s s q q pd d
p p df f
u R i L i L i u
u u L i
( )
( )
δ ϕ
δ ϕ
⎧ = +⎪⎨
= +⎪⎩
ˆ sin
ˆ cos
sd
q s
i i
i i
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Electromagnetic torque
( ) ( )ψ ψ ⎡ ⎤= − = − +⎣ ⎦3 3
2 2e q q q q qd d d d df f T p i i p L L i i L i i
Neglecting the stator resistance
δ ω
δ ω
= = −⎧⎪⎨
= = +⎪⎩
sin
ˆcos
s q qd
q s pd d
u u L i
u u L i u
δ
ω
δ
ω
−⎧=⎪
⎪⎨⎪ = −
⎪⎩
ˆcos
sin
s pd
d
sq
q
u ui
L
ui
L
=>
( )( )δ δ δ
ω ω
δ δ
ω
⎡ ⎤− −⎢ ⎥= − −
⎢ ⎥⎣ ⎦
⎡ ⎤⎛ ⎞= − + −⎢ ⎥⎜ ⎟
⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
2 2
2
ˆ ˆ ˆ ˆ ˆcos sin sin3
2
ˆ ˆ ˆ3 1 1 sin sin 2
2 2
q s p sd p se
q qd
s p s
qd d
L L u u u u uT p
L L L
u u p u
X X X
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Operator inductances of synchronous machine
Laplace transformation of the voltage and flux-linkage equations
ψ ψ ωψ
ψ ψ
ψ
ψ
ψ ψ ωψ
ψ ψ
⎧ ⎛ ⎞= + − −⎪ ⎜ ⎟
⎝ ⎠⎪⎪ ⎛ ⎞
= + −⎪ ⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞
= + −⎨ ⎜ ⎟⎝ ⎠⎪⎪ ⎛ ⎞
= + − +⎪ ⎜ ⎟⎝ ⎠⎪
⎪⎛ ⎞⎪ = + −⎜ ⎟
⎪ ⎝ ⎠⎩
0
0
0
0
0
0
0
ds qd d d
f f f f f
DD D D
qq s q q d
QQ Q Q
u R i s
s
u R i ss
R i s s
u R i ss
R i ss
ψ
ψ
ψ
ψ
ψ
⎧ = + +⎪⎪ = + +⎪⎪
= + +⎨⎪= +⎪
⎪= +⎪⎩
d d d df f dD D
f df d f f fD D
D dD d fD f D D
q q q qQ Q
Q qQ q Q Q
L i L i L i
L i L i L i
L i L i L i
L i L i
L i L i
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ψ ψ
ψ ψ
ψ ψ
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 0
0 0
0 0
d dd d
d df dD f f
f df f fD f
DdD fDD DD D
ii
s sL L Li
L L L is s
L L L ii
s s
ψ ψ
ψ ψ
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 0
0 0
q qq q
q qQ
qQ QQ QQ Q
iiL Ls sL L i
is s
The flux differences expressed using the current differences
Typical initial values: = = =0
0 0 00; f
D Q f
f
ui i i
R
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Quadrature-axis operator inductance
ψ ψ ψ ψ
⎛ ⎞= + − − = −⎜ ⎟
⎝ ⎠
0 00 =>
Q Q Q QQ Q Q Q
R iR i s
s s s
ψ ψ
⎛ ⎞ ⎛ ⎞⎛ ⎞− = − + = − −⎜ ⎟ ⎜ ⎟⎜ ⎟ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 00 =>
q qQ qQQ qQ q Q Q Q q
Q Q
i L s iL i L i i i
s s R L s s
( )ψ
ψ ⎛ ⎞⎛ ⎞ ⎛ ⎞
− = − − = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ ⎝ ⎠ ⎝ ⎠⎝ ⎠
20 0 0q qQ q q
q q q q q
Q Q
L s i iL i L s i
s R L s s s
Voltage equation for the damper winding
Flux-linkage equation for the damper winding
ψ
ψ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
− = − + = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0 0 0 0q q q qQ q
q q q qQ Q q q qQ qQ Q
i i L s i
L i L i L i L is s s R L s s
=>
Flux-linkage equation for the stator winding
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Quadrature-axis subtransient inductance II
( ) σ →∞ →∞
⎛ ⎞= = − = − =⎜ ⎟
⎜ ⎟+
⎝ ⎠
2 2'' lim lim
qQ qQq q q q qQ q
s s
Q Q Q
L s LL L s L L L
R L s L
At the beginning of a transient process
σ σ σ σ
σ
= + = ++ +'' 11 1
Q mqq s sQ mq
Q mq
L LL L LL L
L L
L’’q is called quadrature-axis subtransient inductance
( ) = −+
2qQ
q qQ Q
L sL s L
R L s
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Quadrature-axis operator inductance III
( )
σ
+ −
= − = − =+
+ +
⎛ ⎞⎜ ⎟+ − +⎜ ⎟ +⎝ ⎠= = =
++ +
+
=+
2 2
2
2
''
''0
''''
''0
1 1
1
1 11 1
1
1
qQ qQQ Qq q
qQ Q Q Q Qq q q
Q QQ Q
Q Q
qQ Q Qq q qQ
Q Q qQq q
Q Q q
Q Q
qq
q
L LL Ls L L s s
L s R R L RL s L L
L LR L ss s
R R
L L LL L s s
L R T sRL L
L L T ss sR R
sT
Ls
T
where σ = = = =
'''' '' '' ''0 0 0 0;
qQq Q q qQ Q Q
Q q
LLT T T T T
R L
open-circuit short-circuittime constant time constant
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Quadrature-axis operator inductance IV
( )
+
⎛ ⎞= = + = + −⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
''0
'' ''
'' '' ''
1
1 1 1 1 1 1
1 1 1
q
q q qq q
q q q
s
T A BsL s s sL LL L
s s s sT T T
An equation needed later for the inverse Laplace transformation
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Short circuit at the terminals of a PM synchronous motor
B
H
r B
cH
Magnetic characteristicof the permanent magnets
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-1000
-800
-600
-400
-200
0
200
400600
800
1000
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
L i n e v o l t a g e s [ V ]
-15000
-10000
-5000
0
5000
10000
15000
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
L i n e c
u r r e n t s [ A ]
Line voltage and current in a 3-phase short circuit
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-350000
-300000-250000
-200000
-150000
-100000
-500000
50000
100000
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
T o r q u e [ N m ]
Torque and minimum flux density in permanent magnets
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
M i n i m u m
f l u x d e n s i t y [ T ]
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-12000
-10000
-8000
-6000
-4000
-2000
0
2000
4000
6000
800010000
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
L i n e c
u r r e n t s [ A ]
-1000
-800
-600
-400
-200
0
200
400600
800
1000
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
L i n e v o l t a g e s [ V ]
Line voltage and current in a 2-phase short circuit
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-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
M i n i m u m
f l u x d e n s i t y [ T ]
-150000
-100000
-50000
0
50000
100000
150000
0 50 100 150 200 250 300 350 400 450 500
Time [ms]
T o r q u e [ N m ]
Torque and minimum flux density in permanent magnets
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Direct-axis operator inductance
Laplace transformed voltage and flux-linkage equations
ψ ψ ωψ
ψ ψ
ψ ψ
ψ ψ ωψ
ψ
ψ
⎧ ⎛ ⎞= + − −⎪ ⎜ ⎟⎝ ⎠⎪
⎪ ⎛ ⎞= + −⎪ ⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= + −⎨ ⎜ ⎟
⎝ ⎠⎪⎪ ⎛ ⎞
= + − +⎪ ⎜ ⎟⎝ ⎠⎪
⎪ ⎛ ⎞⎪ = + −⎜ ⎟⎪ ⎝ ⎠⎩
0
0
0
0
0
0
0
ds qd d d
f f f f f
DD D D
qq s q q d
Q
Q Q Q
u R i ss
u R i ss
R i ss
u R i ss
R i s s
ψ ψ
ψ ψ
ψ ψ
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 0
0 0
0 0
d dd d
d df dD
f f f df f fD f
DdD fDD DD D
ii
s sL L L
iL L L is s
L L L ii
s s
ψ ψ
ψ ψ
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤
=⎢ ⎥ ⎢ ⎥⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 0
0 0
q qq q
q qQ
qQ QQ QQ Q
iiL Ls s
L L iis s
d-axis flux and field winding current will be solved as functions of
the d-axis current and field winding voltage.
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Direct-axis operator inductance II
( ) ( )
( ) ( )
ψ
ψ
⎛ ⎞⎛ ⎞
− = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
00 0
0 00
+
+
f d d
d d d d f
f f d f d d d f
ui
L s i G s us s s
i uii sG s i F s u
s s s
After eliminating the damper-winding current and field-winding flux linkage,
we get
Ld(s) is the direct-axis operator inductance
( ) ( )
⎧ ⎡ ⎤⎛ ⎞⎛ ⎞⎪⎢ ⎥⎜ ⎟= + − + −⎨ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎪ ⎝ ⎠⎣ ⎦⎩
⎫⎛ ⎞⎛ ⎞+ − ⎪⎜ ⎟⎜ ⎟+ − − ⎬⎜ ⎟⎜ ⎟
− ⎪⎝ ⎠⎝ ⎠⎭
22
0 0
2 2 22
00 2
1 1 1
2 1
df dD Dd d d f D d f d
D fD dD f df df fD dDD f d
D f D f fD
LLL s L sL T T D s L L L L
L L L L L L L Ls T T L
L L L L L
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Direct-axis operator inductance and time constants
The denominator D(s) is
Using time constants, the direct-axis operator inductance can be written
( )( )( )
( )
( )( )
µ +=+ +
+=
+ +
00 ' ''
0 0
0 0
' ''0 0
1
1 1
1
1 1
df D Dd f
f d d
f Dd
f d d
L T sG s T L T s T s
T T sF s
L T s T s
( ) ( ) ⎛ ⎞
⎜ ⎟= + + + −⎜ ⎟⎝ ⎠
22
0 00 01 1 fD
D D f f D f
LD s s T T s T T
L L
( ) ( )( )
( )( )
+ +=
+ +
' ''
' ''0 0
1 1
1 1
d dd d
d d
T s T sL s L
T s T s
The other coefficient operators in the ψ d, i f equations are
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Time constants( )
( )
( )
( )( )
( )( )
+ += =
+ +
' ''
' ''
0 0
1 1
1 1
d dd d
d d
T s T sN sL s L
D s T s T s
σ σ σ
− = ≈ + ≈ =
− = ≈ = =+
'00 0 0
1
00''00
2 00
1
1
f Dd f f
f
D f DDd fD fD fD
D D f
LT T T T
s R
T T LT T
s T T RSubtransient open-circuittime constant
Transient open-circuittime constant
Roots of the denominator
( ) ( ) ⎛ ⎞
⎜ ⎟= + + + − =⎜ ⎟⎝ ⎠
2
0 00 02 2
1 11 0
fDD D f f
D f
LD sT T T T
s L Ls s
( )
σ
⎡ ⎤+ ⎢ ⎥− = = ± −
⎢ ⎥+
⎢ ⎥⎣ ⎦
0 00 0
2
00
11 1 4
2
D D f f fD
D f
T T T T T
s
T T
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Time constants
Roots of the numerator
σ σ σ σ
σ σ σ
σ σ σ σ σ
≈ + ≈ =
≈ ≈ ≈ =+
'0 0 0
''
00 '' '' '' ''0'' 0
0 0
f Dd dD df f df f df
f
dD fD f
D fD fDd d d d d Dd
D DdD df f d df d df d df
LT T T T
RL
T T T L L L T L L
T T T L L L R
Subtransientshort-circuittime constant
Transientshort-circuit
time constant
( ) ( )
( )
( )( )
( )( )
+ += =
+ +
' ''
' ''
0 0
1 1
1 1
d dd d
d d
T s T sN sL s L
D s T s T s
( ) ( )σ σ σ = + + + =''
0 00 02 2
1 10dD DdD df f fD f
dd
N s LT T T T
s Ls L s
( )
σ σ σ
σ σ
⎡ ⎤⎢ ⎥
+ ⎢ ⎥− = = ± −⎢ ⎥
+⎢ ⎥⎢ ⎥⎣ ⎦
''
000 0
2
0 0
11 1 4
2
dD fD f
DdD df f d
DdD df f
LT T
T T LT
s
T T
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Direct-axis subtransient and transient inductances
Subtransient inductance is effective at the beginning of a transient process
( )→∞
+ −
= = − =−
2 2 ' ''
'' 2 ' ''0 0
2lim
DdD f df df fD dD d dd d d d
s D f fD d d
L L L L L L L T T L L s L L
L L L T T
σ −
= ≈ =⎛ ⎞
+ − −⎜ ⎟⎝ ⎠
' '' ''
'' '' '' 0
0 0 ''1
d d dd d d df d
ddd d d
d
T T T L L L L
T L
T T T L
Transient inductance
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( )
⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= = + +
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟
⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
' ''0 0
''
' '' ' ''
' '' '
' ''
1 1
1 1
1 1 1 1
1 1 1 1 1 1 1
1 1
d d
d d
d d d d
d dd d d
d d
s sT T A B C
sL s sLs s s s s
T T T T
sL LL L Ls sT T
An equation (Heaviside expansion) needed later for the
inverse Laplace transformation
Direct-axis operator inductance
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Typical value s
Turbo-generators(with solid
rotor)
Salient-pole generatorswith damper winding
p < 8 p > 8
Salient-pole generatorswithout damper winding
p < 8 p > 8
Salient-polemotors
with damper
winding
xd (p.u.) 1.5–2.5 0.95–1.78 0.83–1.6 0.98–1.7 0.86–1.5 0.85–2.5
x 'd (p.u.) 0.15–0.35 0.15–0.37 0.23–0.34 0.2–0.35 0.25–0.4 0.22–0.56
x"d
(p.u.) 0.1–0.25 0.08–0.24 0.16–0.24 0.2–0.35 0.25–0.4 0.11–0.32
xq (p.u.) 1.2–2.3 0.46–0.91 0.57–0.89 0.52–0.9 0.45–0.8 0.5–1.5
x"q (p.u.) 0.1–0.25 0.08–0.26 0.17–0.25 0.52–0.9 0.45–0.8 0.11–0.32
T 'd0 (s) 5–15 2–10 4.2–10 2–10 1.5–8 1–7
T 'd (s) 0.6–2.0 0.4–2.5 1.0–2.0 0.5–2.5 0.55–2.5 0.2–1.5
T"d (s) 0.02–0.6 0.02–0.08 0.02–0.05 – – 0.004–0.06
Ta (s) 0.1–0.7 0.04–0.25 0.07–0.15 0.09–0.6 0.1–0.6 0.02–0.15
Reactances and time constants
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Sudden three-phase short circuit
Currentmeasurement
Torquemeasurement
Grid
uf
if
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Sudden three-phase short circuit
Calculation of short-circuit currents is needed for
• design and protection of electrical networks
• dimensioning of switch-gear
• design of electrical machines
Calculation of dynamic forces and torques is needed for
• mechanical dimensioning of the machine, its coupling andfoundations
Two-phase short circuits occur more often than three-
phase short circuits.
Three-phase short-circuit tests are used for determining
machine parameters.
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No-load conditions before the short circuit
Three-phase short-circuit at t =0:
Speed of the rotor is assumed to be constant
Basic conditions
ωψ ω
ψ
= = = =
= = == =
0 0 0 0
0 0 0 0
0 0
0
ˆ0; 0
d q D Q
s q d df f
q d
i i i i
u u L iu
= =
= 0
0d q
f f f
u u
u R i
E ti f t i t l i
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Equations for transient analysis
Laplace transformed voltage and flux-linkage equations
ψ ψ ωψ
ψ ψ
ψ ψ
ψ ψ ωψ
ψ ψ
⎧ ⎛ ⎞= + − −⎪ ⎜ ⎟⎝ ⎠
⎪⎪ ⎛ ⎞= + −⎪ ⎜ ⎟
⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= + −⎨ ⎜ ⎟⎝ ⎠⎪⎪ ⎛ ⎞
= + − +⎪ ⎜ ⎟⎝ ⎠⎪
⎪
⎛ ⎞⎪ = + −⎜ ⎟⎪ ⎝ ⎠⎩
0
0
0
0
0
0
0
ds qd d d
f f f f f
D
D D D
qq s q q d
QQ Q Q
u R i s
s
u R i ss
R i s s
u R i ss
R i ss
ψ ψ
ψ ψ
ψ ψ
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥
⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− = −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 0
0 0
0 0
d dd d
d df dD f f
f df f fD f
DdD fDD DD D
ii
s sL L L iL L L i
s sL L L i
i
s s
ψ ψ
ψ ψ
⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤
=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
0 0
0 0
q qq q
q qQ
qQ QQ QQ Q
iiL Ls s
L L ii
s s
Equations II
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( ) ( )
( )
( ) ( )
ψ ψ
ψ ψ
⎛ ⎞⎛ ⎞− = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞− = −⎜ ⎟
⎝ ⎠⎛ ⎞⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
00 0
0 0
0 00
+
+
f d dd d d d f
q qq q q
f f d f d d d f
uiL s i G s u
s s s
iL s i
s si ui
i sG s i F s us s s
Equations II
Stator flux-linkages and excitation current as functions of stator currents
and excitation voltage
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Modelling of the short circuit
Laplace transformed voltage and flux-linkage equations for
stator winding after substituting the zero initial values
ψ
ψ ωψ
ψ ωψ
⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎨
⎪ = + +⎩
0
0
0
d
s qd d
s q q d
R i s sR i s
( )( )
ψ
ψ
ψ
⎧ − =⎪⎨⎪ =⎩
0
d
d d d
q q q
L s isL s i
After substituting the fluxes
( )[ ] ( )
( ) ( )
ω
ω
⎧ + − =
⎪⎨ ⎡ ⎤+ + = −⎪ ⎣ ⎦⎩0
0
ˆ
s q qd d
ss q qd d
R sL s i L s i
uL s i R sL s i
s
S l ti f th t t t
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Solution for the stator currents
whereD(s)
is the “characteristic polynomial”
( )
( )
( )
( )
ω ⎧= −⎪
⎪⎨
+⎪
= −⎪⎩
0
0
ˆ
ˆ
qsd
s ds
q
L sui
s D s
R sL su
i s D s
( ) ( )[ ] ( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )
ω
ω
⎡ ⎤= + + +⎣ ⎦
⎧ ⎫⎡ ⎤⎪ ⎪= + + + +⎨ ⎬⎢ ⎥
⎣ ⎦⎪ ⎪⎩ ⎭
2
22 21 1
s s q qd d
sq sd
qd d d
D s R sL s R sL s L s L s
RL s L s s sR
L s L s L s L s
Si lifi ti ( ) ( )'' ''li liL L L L
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Simplifications
where = =⎡ ⎤ ⎡ ⎤+⎣ ⎦+⎢ ⎥
⎢ ⎥⎣ ⎦
'' ''
'' ''
'' ''
22
1 1
qda
s qds
qd
L LT
R L LR
L L
( ) ( ) ( )ω ω ⎧ ⎫⎛ ⎞⎪ ⎪
qd
aa
D s L s L s sT T
( )
ω
ω
= −⎡ ⎤⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
02
2
ˆ1
sd
da
ui
sL s sT
( ) ω
= −⎡ ⎤⎛ ⎞⎢ ⎥+ +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
02
2
ˆ1
sq
qa
ui
L s sT
( ) ( ) ( ) ω ⎧ ⎫
≈ + +⎨ ⎬⎩ ⎭
2 22qd
a
D s L s L s s sT
Equations for the currents
is the armature short-circuit time constant
( ) ( )→∞ →∞
= =lim , limq qd ds s
L L s L L s
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Inverse Laplace transform for the currents
where
Inverse transformation of a product of two functions is obtained
using the convolution integral
( ) ( ){ } ( ) ( )−
⋅ = ∗ 11 2 1 2L f s f s f t f t
( ) ( ) ( ) ( ) ( ) ( )τ τ τ τ ∗ = − = −∫ ∫1 2 1 2 1 20 0d d
t t
f t f t f f t t f t f t
( ){ }( )
ω
ω
− − −
⎧ ⎫⎪ ⎪⎪ ⎪⎧ ⎫ ⎪ ⎪
= − ∗⎨ ⎬ ⎨ ⎬⎡ ⎤⎩ ⎭ ⎛ ⎞⎪ ⎪⎢ ⎥+ +⎜ ⎟⎪ ⎪⎢ ⎥⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭
1 1 10 2
2
1 1ˆL L L
1d s
d
a
i s usL s
sT
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Inverse Laplace transform for the currents II
ω
− − −⎡ ⎤⎛ ⎞⎛ ⎞
= − + − + − −⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
' ''
0 '' ' ''
1 1 1 1 1 1
ˆ e e e cos'ad dt T t T t T
sd d d d d d di u tX X X X X X
ω −= − 0''
ˆe sinat T sq
q
ui t
X
Current in stator phase a
ϑ ϑ ϑ ω ϑ = − = + 0cos sin ;sa r q r r r di i i t
After the convolution integration, the currents on the d- and q-axis
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Current in phase a
Substituting the rotor angle gives
( )ω ϑ
ϑ
− −
− −
⎧⎡ ⎤⎪⎢ ⎥⎛ ⎞⎛ ⎞⎪
= − + − + − +⎢ ⎥⎨ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎪⎢ ⎥⎪⎣ ⎦⎩
⎛ ⎞ ⎛ ⎞− + − −⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
' ''
0 0'' '
III III
0'' '' '' ''
IV
1 1 1 1 1ˆ e e cos
'
1 1 1 1 1 1 e cos e
2 2
d d
a
t T t T sa s r
d d d d d
t T t T r
q qd d
i u tX X X X X
X X X X ( )ω ϑ
⎫⎪⎪
+ ⎬⎪
⎪⎭
0
V
cos 2a r t
ω ϑ
ω ϑ
− − −
−
⎡ ⎤⎛ ⎞⎛ ⎞= − + − + − −⎢ ⎥⎜ ⎟⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
+
' ''
0 '' ' ''
0''
1 1 1 1 1 1ˆ e e e cos cos
'
ˆ
e sin sin
ad d
a
t T t T t T sa s r
d d d d d d
t T sr
q
i u tX X X X X X
u
tX
ϑ ω ϑ = + 0r r t
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Current in phase a
The components of the phase current
I steady-state short-circuit current
II transient component that is damped by the time constant
III subtransient component that is damped by the time constant
IV DC component that is damped by the time constant T a
V double-frequency component that is damped by the time constant T a
'dT
''
dT
Components of a phase current
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0.00 0.10 0.20 0.30 0.40 0.50
Time [s]
C
u r r e n t
p p
I II
IV
VIII
Subtransient period
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p
0.00 0.01 0.01 0.02 0.02
Time [s]
C
u r r e n t
I II
IV
VIII
Three phase short circuit; field current
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Three-phase short circuit; field current
( )− = −0
f
f d d
ii sG s i
s
ω − −⎡ ⎤⎛ ⎞ ⎛ ⎞
= − − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
'/ /0 ' '
1+ 1 e 1 e cosadt T t T d d f f d d
X X i i t
X X
Substituting the d-axis current and doing the inverse transformation,
we obtain
Three-phase short circuit at the terminals
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Before the short circuit, the generator
is operating at its rated operation point.
S = 8.2 MVA
cosfii = 0.8 capacitiveUs = 6300 V
f = 50 Hz
Uf = 150 V
A three-phase short-circuit occurs at
t =20 ms.
=> Us = 0
Uf = 150 V
of a 8.2 MVA synchronous generator
After the short circuit, the shaft torque is kept constant. The speed of
the rotor is defined by the equation of motion.
10000
Three-phase short circuit; 8.2 MVA generator
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-10000
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
10000
0 20 40 60 80 100 120 140 160
Time [m s]
L i n e v o
l t a g e s
[ V ]
-15000
-10000
-5000
0
5000
10000
15000
20000
0 20 40 60 80 100 120 140 160
Time [ms]
L i n e
c u r r e n
t s [ A ]
Three-phase short circuit; 8.2 MVA generator
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0
200
400
600
800
1000
1200
1400
1600
1800
0 20 40 60 80 100 120 140 160
Time [ms]
F i e l d c u r r e n t
[ A ]
p ; g
600000
Three-phase short circuit; 8.2 MW generator
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-1000000
-800000
-600000
-400000
-200000
0
200000
400000
600000
0 20 40 60 80 100 120 140 160
Time [ms]
T o r q u e [ N m ]
74
75
76
77
78
79
80
81
82
8384
0 20 40 60 80 100 120 140 160
Time [ms]
S p e
e d [ r a d / s ]
Two-phase short circuit at the terminals of
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a 8.2 MVA synchronous generator
A two-phase short circuit is most
severe, if it occurs at a time instant
when the voltage between the
shorted lines is zero. At this instant,
the stator flux associated with the
shorted branch has a maximum
value. In the short circuit, the flux
“freezes” to this maximum value
and the rotor rotates in a static,slowly decaying flux component.
Two-phase short circuit; 8.2 MVA generator
10000
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-10000
-8000
-6000
-4000
-2000
0
2000
40006000
8000
10000
0 20 40 60 80 100 120 140 160
Time [ms]
L
i n e v o
l t a g e s
[ V ]
-20000
-15000
-10000
-5000
0
5000
10000
15000
20000
0 20 40 60 80 100 120 140 160
Time [m s]
L i n e c u r r e n
t s [ A ]
Two phase short circuit; 8 2 MVA generator
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Two-phase short circuit; 8.2 MVA generator
0
200
400
600
800
1000
1200
1400
1600
0 20 40 60 80 100 120 140 160
Time [ms]
F i e l d c u r r e n
t [ A ]
1000000
Two-phase short circuit; 8.2 MVA generator
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-1500000
-1000000
-500000
0
500000
0 20 40 60 80 100 120 140 160
Time [m s]
A i r - g a p t o r q u e [ N m ]
73
74
75
7677
78
79
80
81
82
83
0 20 40 60 80 100 120 140 160
Time [ms]
R o t a t i o n s p e e d [ r a d / s ]
Asynchronous operation of h hi
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synchronous machines
( )ω
ω = − ≠1 ; 0s M S S p
Currentmeasurement
Torquemeasurement
Grid
u Rif f=
if
R
Asynchronous operation of a synchronous machine
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y p y
Asynchronous operation
• After a drop out of synchronism caused, for instance,by an over load or loss of excitation
• A self-starting synchronous motor when starting
Typically, the field winding is either short-circuited
or connected in series with a resistor.
= 0 f u
Stator voltage in rotor frame of reference
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( )ω ϕ += j
ˆ e s uts
s su u
( )γ ω γ = − + 01 sS t
( )ϕ γ ω −= 0 j jˆ e eu sr S ts su u
( )
( )
ω ϕ γ
ω ϕ γ
= + −⎧⎪⎨
= + −⎪⎩
0
0
ˆ cos
ˆ sin
d s s u
q s s u
u u S t
u u S t
( )ϕ γ ω −= 0 j jˆ e eu sS tsu u=⎧
⎨ = −⎩ jd
q
u u
u u
Constant slip
Stator voltage in stator and rotor frames of reference
d- and q-components
Complex notation (phasors)
Laplace transformed voltage equations andthe definition of the operator inductances
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the definition of the operator inductances
ψ ψ ωψ
ψ ψ ωψ
⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎪⎨
⎛ ⎞⎪ = + − +⎜ ⎟⎪ ⎝ ⎠⎩
0
0
ds qd d d
q
q s q q d
u R i ss
u R i s s
( ) ( )
( )
ψ ψ
ψ ψ
⎧ ⎛ ⎞⎛ ⎞− = − −⎪ ⎜ ⎟ ⎜ ⎟
⎝ ⎠⎪ ⎝ ⎠⎨⎛ ⎞⎪ − = −⎜ ⎟⎪⎝ ⎠⎩
00 0
0 0
+ f d d
d d d d f
q qq q q
uiL s i G s u
s s si
L s is s
The initial values of the flux linkages are put to zero, and theLaplace variable s is replaced by jSω s. Field winding is assumedto be short-circuited, u f = 0
Solution of stator fluxes (Rs = 0)
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( )
( )
ω ψ ω ψ
ω ψ ω ψ
= = + − −⎧⎪⎨
= − = + + −⎪⎩
j 1
j j 1
d s d s d s q
q s q s q s d
u u R i S S
u u R i S S
≈ 0sR
Voltage equations
Solution of flux linkages after neglecting the stator resistance
( )
( )
ω ψ ω ψ
ω ψ ω ψ
= − −⎧⎪⎨
= − −⎪⎩
j 1
j 1
s s qd
s s qd
u S S
u S S =>
( )
( )
( )
( )
ω ψ
ω ω
ω ψ
ω ω
−⎧= = −⎪ −⎪
⎨ − −⎪ = = −⎪ −⎩
2
2
1 2 j
j 1 2
j 1 2
j 1 2
sd
ss
sq
ss
S uu
S
S uu
S
( )
ψ ω ϕ γ ω
ψ ω ϕ γ ω
π⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎪⎨⎪ = − + −⎪⎩
0
0
ˆ cos2
ˆsin
s s uds
sq s u
s
u S t
uS t
Instantaneous flux linkages
( )
( )
ω ϕ γ
ω ϕ γ
= + −⎧⎪⎨
= + −⎪⎩
0
0
ˆ cos
ˆ sin
d s s u
q s s u
u u S t
u u S t =>
Solution of stator currentsC t t f th fl li k
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( ) ( )
( ) ( )
ϕ
ϕ
ω ω ω ω
ω ω ω ω
π⎛ ⎞− +⎜ ⎟⎝ ⎠
−
⎧⎪ = − =⎪⎨⎪ = − = −⎪⎩
j2
j
je
j j
e j j
d
q
ds d s s d s
q s q s s q s
u ui
L S L S
u u
i L S L S
( )
( ) ( )
ω ϕ ϕ γ ω ω
ω ϕ ϕ γ
ω ω
π⎧ ⎛ ⎞= + − − −⎜ ⎟⎪ ⎝ ⎠⎪
⎨⎪ = − + − −
⎪⎩
0
0
ˆcos
2 j
ˆcos
j
sd s u d
s d s
sq s u q
s q s
ui S t
L S
ui S t
L S
( )
( )
ψ ω
ψ ω
⎧ =⎪⎨
=⎪⎩
j
j
d d s d
q q s q
L S i
L S i =>
Where are the arguments of the operator inductances.
Instantaneous currents
Current components from the flux linkages
A phase current
γ γ = −cos sinsa d qi i i
ϕ ϕ qd and
Phase current
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( ) ( )
( ) ( ) ( )
ω ϕ ϕ γ ω γ ω ω
ω ϕ ϕ γ ω γ
ω ω
π⎛ ⎞
= + − − − − +⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠
π⎡ ⎤+ + − − − + −⎢ ⎥⎣ ⎦
0 0
0 0
ˆ
cos cos 12 j
ˆ cos cos 1
2 j
ssa s u d s
s d s
ss u q s
s q s
ui S t S tL S
uS t S t
L S
( ) ( )⎡ ⎤= − + +⎣ ⎦1
cos cos cos cos2
x y x y x y
( ) ( )
( )
( )
ω ϕ ϕ ω γ ϕ ϕ ω ω
ω ϕ ϕ ω γ ϕ ϕ
ω ω
π π⎧ ⎫⎛ ⎞ ⎡ ⎤= + − − + − + − + +⎨ ⎬⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎩ ⎭
π π⎧ ⎫⎛ ⎞ ⎡ ⎤+ + − − + − + − + −⎨ ⎬⎜ ⎟ ⎢ ⎥
⎝ ⎠ ⎣ ⎦⎩ ⎭
0
0
ˆcos cos 1 2 2
2 22 j
ˆ cos cos 1 2 2
2 22 j
ssa s u d s u d
s d s
ss u q s u q
s q s
ui t S t
L S
ut S t
L S
Modification using the identity
The phase current
is of the form ( ) ( )ω ϕ ω ϕ = + + − +⎡ ⎤⎣ ⎦1 1 2 2ˆ ˆcos cos 1 2sa sa s sa si i t i S t
A closer study of the phase current components
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( ) ( )
ϕ ω
ω ω ω
π⎛ ⎞−⎜ ⎟⎝ ⎠
⎡ ⎤= +⎢ ⎥
⎢ ⎥⎣ ⎦
j j2
1
ˆ 1 1e e
2 j j
usts
sas s q sd
ui
L S L S
( ) ( )( )
ϕ γ ω
ω ω ω
π⎛ ⎞− −⎜ ⎟ − −⎝ ⎠⎡ ⎤
= −⎢ ⎥⎢ ⎥⎣ ⎦
0 j 2* j 1 22 2
ˆ 1 1e e
2 j j
usS ts
sas s q sd
ui
L S L S
( ) ( ) ( )
( ) ( ) ( )[ ]
ω ϕ ω ω ω
ω ϕ ω ω ω
= + +
+ − − +
1
2
ˆ 1 1cos2 j j
ˆ 1 1 cos 1 2
2 j j
ssa ss s q sd
ss
s s q sd
ui tL S L S
uS t
L S L S
Fundamental frequency component in phasor notation
Complex conjugate of the other current phasor
Phase current
Electromagnetic torque
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( )ψ ψ = −32e q qd dT p i i
Flux linkages
Currents
( )
ψ ω ϕ γ ω
ψ ω ϕ γ ω
π⎧ ⎛ ⎞= + − −⎜ ⎟⎪ ⎝ ⎠⎪⎨⎪ = − + −⎪⎩
0
0
ˆ cos2
ˆsin
ss ud
s
sq s u
s
u S t
uS t
( )
( ) ( )
ω ϕ ϕ γ ω ω
ω ϕ ϕ γ
ω ω
π⎧ ⎛ ⎞= + − − −⎜ ⎟⎪ ⎝ ⎠⎪⎨⎪ = − + − −⎪⎩
0
0
ˆcos
2 j
ˆcos
j
s
d s u ds d s
sq s u q
s q s
ui S t
L S
ui S t
L S
Electromagnetic torque II
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( )
( )
( )
ψ ψ
ϕ ω ϕ γ ϕ
ω ω
ϕ ω ϕ γ ϕ
ω ω
= −
π π⎧ ⎫⎛ ⎞ ⎡ ⎤= + + + − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎩ ⎭
π π⎧ ⎫⎛ ⎞ ⎡ ⎤− − + + − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎩ ⎭
2
02
2
02
3
2
ˆ3 cos cos 2 2 2
2 24 j
ˆ3 cos cos 2 2 2
2 24 j
e q qd d
ss ud d
s sd
sq s u q
s q s
T p i i
puS t
L S
puS t
L S
( ) ( ) ( ) ( )
ϕ ϕ
ω ω ω ω ω ω
π π⎡ ⎤⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎡ ⎤⎢ ⎥⎝ ⎠ ⎝ ⎠= + = +⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦
⎢ ⎥⎣ ⎦
2 2
1 2 2
cos cosˆ ˆ3 3 1 12 2
Im j j j4 4 j
qds s
es q sdsds sq s
pu puT
L S L SL S L S
( )ω ϕ = + +1 2ˆ cos 2e e e s T T T T S t
The torque is of the form
The average torque is
Operator inductances and slip frequency
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( )
⎛ ⎞= + −⎜ ⎟
⎜ ⎟ ⎛ ⎞⎝ ⎠ +⎜ ⎟⎜ ⎟⎝ ⎠
''
''
1 1 1 1 1
1q q qq
q
sL s sL LLs
T
( )
⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟
⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
' '' '
' ''
1 1 1 1 1 1 1 1
1 1d d dd d d
d d
sL s sL LL L Ls s
T T
( )
ω ω
ω ω ω
⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟
⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
' '' '
' ''
j j1 1 1 1 1 1
j 1 1 j j
s s
sd d dd d d
s sd d
S S
L S L LL L LS S
T T
( )
ω
ω
ω
⎛ ⎞⎜ ⎟= + −⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎜ ⎟+
⎜ ⎟⎝ ⎠
''
''
j1 1 1 1
j 1 j
s
q s q qqs
q
S
L S L LLS
T
Direct axis
Quadrature axis
=>
=>
⎡
The average electromagnetic torque
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ω ω ω
ω ω
ω ω
⎡
⎢ ⎛ ⎞ ⎛ ⎞= − + −⎢ ⎜ ⎟ ⎜ ⎟⎢ ⎝ ⎠ ⎝ ⎠+ +⎢⎣
⎤
⎥⎛ ⎞ ⎥+ −⎜ ⎟⎜ ⎟ ⎥+⎝ ⎠ ⎥
⎦
2 ' ''1 2 ' '' '' ''
' ''
''
'' ''''
ˆ3 1 2 1 21 11 18
1 2 1
1
s d de
ds d d ds sd ds sd d
q
q q s qs q
pu L LT LL L LS T S T
S T S T
L
L L S T S T
= + +
+ + +
1 I II IIII II III
I II III
2 2 2e p p p
p p p
p p p
T T T T S S SS S S
S S S S S S
ω ω ω
= = =I II III' '' ''1 1 1
; ; ; p p p
s d s d s q
S S S
T T T
consists of three terms similar to the steady-state torque of aninduction machine
where
Average torque in asynchronous operation
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a) Total torque;Stator resistance is
taken into account
b) Total torque;
Stator resistanceis neglected
c) Torque caused by the
damper winding
d) Torque caused by thefield winding
Influence of an external resistance connectedin series with the field winding
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Without external resistance With an external resistance
The pulsating torque
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( ) ( )
ϕ γ ω
ω ω ω
π⎛ ⎞− −⎜ ⎟⎝ ⎠
⎡ ⎤= −⎢ ⎥
⎢ ⎥⎣ ⎦
02 j 2 2
j22 2 2
ˆ3 1 1e e
j j4
usS ts
es q sds
puT
L S L S
ω = = −
2
2 2̂3 1 1ˆ0 =>4
se
qds
puS T L L
The reason for the pulsating torque is the saliency of the rotor.
If
At the synchronous speed, this component gives the reluctance torque.
Its peak value is
( ) ( )ω ω = => =2ˆ j j 0s q s edL S L S T
Amplitude of the pulsating torqueas a function of slip
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Fast and slow starting
''
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a) Starting time
b) Starting time ≈ ≈'100 100 sdT
≈ ≈''
10 0.5 sdT
Starting of a synchronous motor
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Currentmeasurement
Torque and speedmeasurement
u Rif f=
if
R
Grid
2.5 MW synchronous motor
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Rated values:
U s = 5500 VI s = 290 A
P = 2.5 MW f s = 50 Hz
When starting:
T shaft = 0Rext = 9 R f
4000
6000Starting of a 2.5 MW synchronous motor
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-8000
-6000
-4000
-2000
0
2000
0 100 200 300 400 500
Time [ms]
L i n e c u r r e n t s [ A ]
-800
-600
-400
-200
0
200
400
600
0 100 200 300 400 500
Time [ms]
F i e
l d c u r r e n t [ A ]
150000
200000
Starting of a 2.5 MW synchronous motor
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-150000
-100000
-50000
0
50000
100000
0 100 200 300 400 500
Time [ms]
A
i r - g a p t o r q u e [ N m ]
0
10
20
30
40
50
60
70
0 100 200 300 400 500
Time [ms]
R o t a t i o
n s p e e d [ r a d / s ]
Starting of a 2.5 MW synchronous motor Space vector of stator current
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-7500
-5000
-2500
0
2500
5000
7500
-7500 -5000 -2500 0 2500 5000 7500
Re [A]
I m [
A ]
Asynchronous operation of a 8.2 MVAsynchronous generator
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Rated values:
U s = 6300 V f s = 50 Hz
I s = 940 AI f = 215 A
S = 8.2 MVA
Under fault:
U f = 0 V T shaft = T N
83
84
85
d / s ]
Asynchronous operation of a synchronous generator
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75
76
77
78
79
80
81
82
0.0 0.5 1.0 1.5 2.0 2.5
Time [ms]
R o t a t i o n s p e e d [ r a d
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
0.0 0.5 1.0 1.5 2.0 2.5
Time [ms]
R o t o r a n g l e [ e l . r a d ]
Time [s]
Asynchronous operation of a synchronous generator
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-300000
-250000
-200000
-150000
-100000
-50000
0
50000
0.0 0.5 1.0 1.5 2.0 2.5
Time [s]
A i r - g a p t o r q u e [ N
m ]
3000
4000
5000
Asynchronous operation of a synchronous generator
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-5000
-4000-3000
-2000
-1000
0
1000
2000
0.0 0.5 1.0 1.5 2.0 2.5
Time [s]
L i n e c u r r e n t [ A ]
-500
-400
-300
-200
-100
0
100
200
300400
500
0.0 0.5 1.0 1.5 2.0 2.5
Time [s]
F i e
l d c u r r e n t [ A ]
Asynchronous operation of a synchronous generator
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Operating characteristics
Terminal voltage 6300. V
Supply frequency 50.0 Hz Terminal current 1652.1 A Peak current 3731.7 A Power factor -0.4784 Ind.
Slip-ring voltage 0.00 V Slip-ring current 196.3 A
Rotational speed 755.3 1/min Air-gap torque -99890 Nm
Equations of synchronous machine usingload angle
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Load angle δ is the angle between the
excitation voltage vector (on q-axis)
and stator voltage vector.
In a balanced stiff three-phase network,the space vector of stator voltage is
The angle of the rotor presented in stator frame of reference is
The angular speed of the rotor, its derivative and slip are
( )ω ϕ += jˆ e s uts su u
γ ω ϕ δ π
= + + −2
s ut
γ δ ω γ δ ω ω δ ω ω
ω ω
−= = + = = = = −
2 2
2 2
d d d d d 1 d; ;
d d d dd ds
ss s
St t t tt t
su
du j quγ
δ
q
d
β
α
Equations of synchronous machine usingload angle
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δ
δ
=⎧⎪⎨ =⎪⎩
ˆ sin
ˆ cos
d s
q s
u u
u u
The equation of motion becomes
and the voltage equations are
The dq-components of the balanced three-phase voltage are
( ) δ
ψ ψ − = +2
2
3 d
2 dd q q d m
J p i i T
p t
ψ δ ω ψ ψ
ψ δ ω ψ ψ
⎧ = + − −⎪⎪⎨⎪ = + + +⎪⎩
d d
d dd d
d d
dd s d s q q
qq s q s d d
u R i
t t
u R it t
Small-signal analysis for synchronous machines
A l t i l hi i ld i l t d t t Th i l
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An electrical machine is seldom in a real steady state. There is alwayssome variation in the shaft torque and supply voltage that causes
disturbances to the operation of the machine. If the variations are small in
amplitude, equations can be linearised to the steady state operation point.
The variables are presented as a sum of a steady state value and
small variation, for instance
The voltage equations for the steady-state variables are
∆ ω ω ∆ω = + = +0 ;d d d si i i
ω ψ
ω ψ
= −⎧⎪⎨
= +⎪⎩
0 0 0
0 0 0
d s d s q
q s q s d
u R i
u R i
Equations for the voltage variations
F th di t b d i bl th lt ti
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For the disturbed variables, the voltage equations are
Subtracting the two sets of equations and assuming that the products
of two small variations can be neglected, we get the equations for
the variations
( ) ( )( )
( ) ( ) ( )
∆ψ ∆ ∆ ω ∆ω ψ ∆ψ
∆ψ
∆ ∆ ω ∆ω ψ ∆ψ
⎧+ = + + − + +⎪⎪
⎨⎪
+ = + + + + +⎪⎩
0 0 0
0 0 0
d
dd
d
dd d s d d s q q
q
q q s q q s d d
u u R i it
u u R i i t
∆ψ ∆ ∆ ω ∆ψ ψ ∆ω
∆ψ ∆ ∆ ω ∆ψ ψ ∆ω
⎧= + − −⎪
⎪⎨⎪ = + + +⎪⎩
0
0
d
dd
d
dd s d s q q
qq s q s d d
u R i
t
u R it
Laplace transformation of the voltage variations
The equations are linear and easy to Laplace transform
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The equations are linear and easy to Laplace transform
The corresponding flux-linkage equations are
Combining the equations, we get
∆ ∆ ∆ψ ω ∆ψ ψ ∆ω
∆ ∆ ∆ψ ω ∆ψ ψ ∆ω
⎧ = + − −⎪⎨
= + + +⎪⎩
0
0
d s d d s q q
q s q q s d d
u R i s
u R i s
( ) ( )
( )
∆ψ ∆ ∆
∆ψ ∆
⎧ = +⎪⎨
=⎪⎩
d d d f
q q q
L s i G s u
L s i
( ) ( ) ( )
( ) ( ) ( )
∆ ∆ ω ∆ ∆ ψ ∆ω
∆ ∆ ω ∆ ω ∆ ψ ∆ω
⎧ = + − + −⎡ ⎤⎣ ⎦⎪⎨
⎡ ⎤= + + + +⎪ ⎣ ⎦⎩
0
0
d s d d s q q f q
q s q q s d d s f d
u R sL s i L s i sG s u
u R sL s i L s i G s u
Variation of the torque ( p=1)
The equation for the varying torque is
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The equation for the varying torque is
The corresponding steady-state torque equation is
Subtracting the two equations, we get
( )ψ ψ − =0 0 0 0 03
2 d q q d mi i T
∆ω ψ ∆ ∆ψ ψ ∆ ∆ψ ∆⎡ ⎤+ − − − =⎣ ⎦0 0 0 03 d2 d
d q q d q d d q mi i i i T J t
( )( ) ( )( ) ∆ω
ψ ∆ψ ∆ ψ ∆ψ ∆ ∆⎡ ⎤+ + − + + = + +⎣ ⎦0 0 0 0 0
3 d
2 dd d q q q q d d m mi i i i J T T
t
Variation of the torque II
The Laplace transformation of the torque variation is
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Substituting the flux-linkage equations, the torque becomes
ψ ∆ ∆ψ ψ ∆ ∆ψ ∆ ∆ω ⎡ ⎤+ − − − =⎣ ⎦ 0 0 0 0
3
2 d q q d q d d q mi i i i T sJ
( ) ( ) ( )
{ }ψ ∆ ψ ∆ ∆ ∆ ∆ω ⎡ ⎤ ⎡ ⎤− − − + − =
⎣ ⎦ ⎣ ⎦
0 0 0 0 0
3
2 d d q q q q d d q f mi L s i i L s i i G s u T sJ
Variation of the voltage components
If the machine is connected to a balanced stiff line voltage, the variation
of the load angle causes a variation in the voltage components
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of the load angle causes a variation in the voltage components
∆ ∆δ δ ∆δ
∆ ∆δ δ ∆δ
= =⎧⎪⎨
= − = −⎪⎩
0 0
0 0
ˆ cos
ˆ sin
d q s
q d s
u u u
u u u
( )∆ δ ∆δ δ ∆δ δ ∆δ
∆δ ∆δ ∆δ
+ = + = +
= + ≈ +
0 0 0 0
0 0 0 0
ˆ ˆ ˆsin sin cos cos sin
cos sin
d d s s s
d q d q
u u u u u
u u u u
( )∆ δ ∆δ δ ∆δ δ ∆δ
∆δ ∆δ ∆δ
+ = + = −
= − ≈ −
0 0 0 0
0 0 0 0
ˆ ˆ ˆcos cos cos sin sin
cos sin
q q s s s
q d q d
u u u u u
u u u u
Forced variations of a synchronous machine
The variations are often periodic. This is the case, for instance, for the
shaft torque of a diesel driven generator For sinusoidal disturbance
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shaft torque of a diesel driven generator. For sinusoidal disturbance
( ) ( )λ ϕ ∆ ∆ ∆ += + = j0 ˆRe where e idt
d d d d di i i i i
( )∆ λ ϕ = + +0 ˆ cosm m m T T T T t
If the sinusoidal disturbance in the torque is small, the machine behaves
linearly and the other parameters will also vary sinusoidally. We can use
phasor variables, for instance
The Laplace transformed equations derived previously can be changesto phasor equations by replacing the Laplace variable s with thefrequency jλ . The variation of the rotation speed can be written
γ δ δ ω ω ∆ω ω ω ∆ω λ∆δ = = + = − = =
d d d => => j
d d ds s
t t t
Voltages and flux linkages
The voltage equations, assuming that u f
is constant and Rs
=0, are
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f
The steady-state flux linkages (Rs=0) and voltages are related as
( ) ( )
( ) ( )
∆ λ λ ∆ ω λ ∆ ψ λ∆δ
∆ λ λ ∆ ω λ ∆ ψ λ∆δ
⎧ = − −⎪⎨
= + +⎪⎩
0
0
j j j j
j j j j
d d d s q q q
q q q s d d d
u L i L i
u L i L i
ψ ω
ψ ω
⎧ =⎪⎪⎨⎪ = −⎪⎩
00
00
qd
s
dq
s
u
u
∆ ∆δ
∆ ∆δ
=⎧⎪⎨
= −⎪⎩
0
0
d q
q d
u u
u u
On the other hand
Current phasor
Combining the three previous equations
( ) ( )∆δ λ λ ∆ λ ∆ λ∆δ⎧ 0j j j jduL i L i
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( ) ( )
( ) ( )
∆δ λ λ ∆ ω λ ∆ λ∆δ ω
∆δ λ λ ∆ ω λ ∆ λ∆δ ω
⎧ = − +⎪⎪⎨⎪− = + +⎪
⎩
00
00
j j j j
j j j j
dq d d s q q
s
qd q q s d d
s
uu L i L i
uu L i L i
( ) ( )
( ) ( )
λ ∆δ λ λ ∆ ω λ ∆
ω
λ ∆δ ω λ ∆ λ λ ∆
ω
⎧⎛ ⎞− = −⎪⎜ ⎟
⎝ ⎠⎪⎨
⎛ ⎞⎪− + =⎜ ⎟⎪
⎝ ⎠⎩
00
00
j j j j
j j +j j
dq d d s q q
s
qd s d d q qs
uu L i L i
uu L i L i
=>
From which we can solve the currents
( )
( )
∆ ∆δ ω λ
∆ ∆δ ω λ
⎧ = −⎪⎪⎨⎪ = −⎪⎩
0
0
j
j
dd
s d
qq
s q
uiL
ui
L
Equation of motion in phasor form
The equation of motion in phasor form is
3
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After substituting the fluxes and currents
( )
( )
∆ ∆δ ω λ
∆ ∆δ ω λ
⎧ =⎪⎪⎨
⎪ = −⎪⎩
0
0
j
j
dd
s d
qq
s q
ui
L
uiL
( ) ( ){ }ψ λ ∆ ψ λ ∆ ∆ λ ∆δ ⎡ ⎤ ⎡ ⎤− − − − = −⎣ ⎦ ⎣ ⎦2
0 0 0 03
j j2
d d q q q q d d mi L i i L i T J
( ) ( ) ∆δ ∆ λ ∆δ
ω ω λ ω λ
⎡ ⎤⎢ ⎥− + − − − = −⎢ ⎥⎣ ⎦
2 20 20
0 0 0 03
2 j j
q dq d d q m
s s q s d
u uu i u i T J
L L
ψ ω
ψ ω
⎧=⎪
⎪⎨
⎪ = −⎪⎩
00
00
qd
s
dq
s
u
u
we get
Equation of motion in phasor form II
The term ( )−
0 0 0 0
3
2 q d d qu i u i
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can be expressed with the aid of the steady-state reactive power
The torque equation becomes
( )
( ) ( )
( ) ( )
( ) ( )
δ δ
δ δ φ δ δ φ
δ φ δ δ φ δ
− = −
⎡ ⎤= + − +⎣ ⎦
= + − +⎡ ⎤⎣ ⎦
0 0 0 0 0 0 0 0
0 0 0 0
0 0 0 0
3 3ˆ ˆcos sin
2 23 ˆ ˆˆ ˆ cos sin sin cos23 ˆˆ sin cos cos sin
2
q d d q s d s q
s s s s
s s
u i u i u i u i
u i u i
u i
( )δ φ δ φ = + − = =⎡ ⎤⎣ ⎦0 0 03 3ˆ ˆˆ ˆ sin sin2 2
s s s su i u i Q
( ) ( )λ ∆δ ∆
ω ω λ ω λ
⎡ ⎤⎢ ⎥+ − − =⎢ ⎥⎣ ⎦
2 202 0 0
2 2
3 3
2 2 j j
q dm
s s q s d
uQ u J T
L L
To shorten the notation, a ‘complex synchronisation coefficient’
is defined
Complex synchronisation coefficient
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( ) ( )( ) ( )
λ λ λ ω ω λ ω λ
⎡ ⎤+ = − − −⎢ ⎥
⎢ ⎥⎣ ⎦
2 20 0
01 3 3
j2 j 2 j
q ds d
s s q s d
u uK K Q
L L
and the equation of motion becomes
( ) ( )λ λ λ λ ∆δ ∆⎡ ⎤− + = −⎣ ⎦2
js d mK J K T
where K d represents damping and K s∆δ is the synchronising torque.
K d
and K s
are frequency-dependent but their values can be
calculated for a certain excitation frequency.
Complex synchronisation coefficient II
( ) ( )δ δ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= − + +⎨ ⎬ ⎨ ⎬
2 22 200 02 2
ˆ ˆ3 1 3 1cos Re sin Res ss Q u uK
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( ) ( )δ δ
ω λ λ ω ω ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ + +⎨ ⎬ ⎨ ⎬
⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭0 02 2
cos Re sin Re2 j 2 j
ss q ds s
QK L L
( ) ( )δ δ λ λ λ ω
⎡ ⎤⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭⎣ ⎦
2
2 20 02
ˆ3 1 1cos Im sin Im2 j j
sdq ds
uK L L
In normal operation,
This means that the properties of the damper winding on the
quadrature axis are important from the point of view of damping.
δ δ δ < >>0 2 2
0 0 030 => cos sin
Numerical study of a synchronous generator using small forced vibrations
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Rated values S = 8.2 MVA
f s = 50 HzU
s
= 6300 VI s = 750 A
cosϕ = 0.80I f = 215 A
Ω Ω Ω λ = +0 ˆ cosm m m t
Forced oscil lation of the mechanical rotation
speed
100
Forced oscil lation of speed and the resultingoscillation in air-gap torque (λ = 50 Hz)
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0
20
40
60
80
300 350 400 450 500
Time [ms]
S p
e e d [ r a d / s ]
-140000
-120000-100000
-80000
-60000
-40000
-20000
0
300 350 400 450 500
Time [ms]
T
o r q u e [ N m ]
Power transfer in the forced oscillation
Variation of speed: λ λ Ω Ω λ =ˆ cos t
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Mass of inertia:
Viscous damping:
Spring constant:
Total torque:
Instantaneous power:
λ
ϕ Ω λ λ = = −
2
J 2
d ˆ sind
T J J tt
λ λ λ
Ω Ω λ λ Ω λ λ
λ = − + +tot
ˆˆ ˆsin cos sinT J t d t k t
λ
λ λ λ λ
Ω
Ω Ω λ Ω λ λ Ω λ λ
λ
=
⎛ ⎞= − + +⎜ ⎟
⎝ ⎠
tot
ˆˆ ˆ ˆ cos sin cos sin
P T
t J t d t k t
λ
ϕ Ω λ = =d
d ˆ cosd
T d d tt
λ Ω
ϕ λ λ = =
k
ˆ
sinT k k t
Average power of the forced oscillation
100
Instantaneous harmonic power (λ = 50 Hz)
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-150
-100
-50
0
50
100
300 350 400 450 500Time [ms]
H a r m o n i c p
o w e r [ k W ]
λ λ λ λ
λ
Ω Ω λ Ω λ λ Ω λ λ
λ
Ω
⎛ ⎞= − + +⎜ ⎟
− ⎝ ⎠
=
∫ave
2
ˆ1 ˆ ˆ ˆcos sin cos sin d
1 ˆ 2
b
a
t
ab t
P t J t d t k t t
t t
d
Is the harmonic power always negative,i.e. is the damping always positive?
If th h i i ti 12
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If the harmonic power is negative,
the machine takes mechanical
power from the shaft forcing the
oscillation. There is damping thattries to reduce the oscillation.
If the harmonic power is positive,
the system tends to increase the
oscillation and the system may beunstable.
From the harmonic power, we can
calculate the damping coefficient
for the system, which should bepositive.
-2
0
2
4
6
8
10
12
0 20 40 60 80 100
Excitation frequency [Hz]
D a m p i n
g c o e f f i c i e n t [ k N m s ]
Damping coefficient after reducing theresistance of the stator winding to zero
12
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-2
0
2
4
6
8
10
0 20 40 60 80 100
Excitation frequency [Hz]
D a m p i n g
c o e f f i c i e n t [ k
N m s ]
Mechanical model for torsional vibrationsof a diesel-generator
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Θ Θ Θ Θ Θ Θ
Θ Θ Θ Θ
+ −−
+ − −
− −+ + +
+ − + − =
2 n n 1 n n 1n nn n n n 12
n n n 1 n 1 n n 1 n
( ) ( )
( ) ( ) ( )
d dd d J B C C dt dt dtdt
K K T t
Equation of motion for a mass of inertia
Synchronousgenerator
Coupling Diesel engine
Torsional vibrations
Θ
2d
Explanations for the terms in the torque balance equation
Inertia torque of mass
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Θ 2
d
dn
n J t
Inertia torque of mass n
Mechanical friction torque acting on mass n
Damping torque component in the shaft
between masses n and n+1
Spring constant of the shaft betweenmasses n and n+1
Active torque on a mass of inertia.Electromagnetic torque in the electrical machine.Piston torque from a cylinder of the diesel engine.
Θ Θ +− 1d( )
dn n
nC t
Θ dd
nnB t
( )nT t
Θ Θ +− 1( )n n nK
80
100
Torque produced by one of the 18 cylinders
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Fundamental frequency is 6.25 Hz
18 cylinders => largest shaft-torque harmonic at 112.5 Hz
-60
-40
-20
0
20
40
60
0.00 0.05 0.10 0.15 0.20
Time [s]
T o r q u e [ k
N m ]
Natural vibration frequencies and modes
2d d
Equation of motion in matrix form
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Θ Θ Θ Θ ⎡ ⎤+ + = ⎣ ⎦
2T
1, 2, 3,2
d d ; ...,
dd N
tt
J C K Τ
Natural vibration modes and frequencies; trial function λ e t θ
λ λ λ λ λ λ λ λ + + + + =2
22
d de e e e e e 0
dd
t t t t t t
ttθ θ θ θ θ θ
J C K J C K
Non-zero solutions for θ possible if λ satisfies the characteristic
equation
( )λ λ + + =2det 0 J C K
λ λ ⎡ ⎤+ + =⎣ ⎦2 0 J C K=>
Natural frequencies and modes
The roots of the characteristic equation are typically complex numbersλ α ω= + ji i i
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ω i is a natural frequency of the mechanical system, α i is associated with
the damping of a vibration mode i. All α i should have negative values,otherwise the system is unstable.
Typically, the damping terms are so small that they can be neglected
if one is only interested in the natural frequencies. In such a case,matrix C is left out of the analysis.
λ α ω = + ji i i
Results from the combined electromechanical model
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Rated values
S = 8.2 MVA f s = 50 Hz
U s = 6300 V
I s = 750 Acosϕ = 0.80I f = 215 A
400
450
500
Torque versus torsion angle characteristicfor the non-linear coupling
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Coupling: Rated torque 110 kNm; Stands 330 kNm
The mass of inertia of the synchronous generator is about twice as
large as the mass of inertia of the diesel engine.
0
50
100
150
200
250
300
350
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14
Torsion angle [rad]
T o r q u e [ k N m ]
Two-phase short circuit in the line
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Current and voltagemeasurement
Torque and speedmeasurement
if
Grid
uf
Z
Diesel engine
0
25005000
7500
10000
o l t a g e
[ V ]
Voltages and currents at the terminals of the machine
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-10000-7500
-5000
-2500
0
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
T e r m i n a l v
-8000
-6000
-4000
-2000
0
2000
4000
6000
8000
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
T e r
m i n a l c u r r e n t [ A ]
600
700
800
900
1000
c u r r e n t [ A ]
Field-winding current and the torques
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0
100200
300
400
500
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
M a g n e t i s a t i o n
-800
-600
-400
-200
0
200400
600
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
T o r q u e [ k N m ]
Air gap Coupling
Synchronisation trial at phase opposition
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Current and voltagemeasurement
Torque and speedmeasurement
if
Grid
uf
Z
Diesel engine
0
25005000
7500
10000
v o l t a g e
[ V ]
Voltages and currents at the terminals of the machine
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-10000-7500
-5000
-2500
0
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
T e r m i n a l v
-20000
-15000
-10000
-5000
0
5000
10000
15000
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
T e r
m i n a l c u r r e n t [ A ]
1000
12001400
16001800
2000
n c u r r e
n t [ A ]
Field-winding current and the torques
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-200
0200
400600
800
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
M a g
n e t i s a t i o n
-1500
-1000
-500
0
500
1000
0 20 40 60 80 100 120 140 160 180 200 220
Time [ms]
T o r q u e [ k N m ]
Air gap Coupling
A complete description of the electromagnetic field requires5 field variables ( E, D, H , B, J ).
Magnetic field analysis for electrical machines
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( , , , , J)
Maxwell’s equations Material conditions
For the frequencies considered,
In addition, we need the boundary conditions for the fields.
ρ ∇ ⋅ =
∇ ⋅ =
∂∇ × = −
∂∂
∇ × = +∂
0
t
t
D
B
B E
D H J
ε
σ
µ
=
=
=
D E
J E
B H
∂
≈∂ 0t
D
Vector potential A and scalar potential φ
∇⎧ H J
∇ × = A B φ ∂
= − − ∇
∂t
A EDefinitions:
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=>
The 5 original field variables and their 15 components have
been compressed to 2 potentials with altogether 4 components.
µ
∇ × =⎧⎪
=⎨
⎪ = ∇ ×⎩
H J
B H
B A µ
⎛ ⎞∇ × ∇ × =
⎜ ⎟⎝ ⎠
1 A J
σ
φ
=⎧⎪
∂⎨= − − ∇⎪ ∂⎩ t
J E
A E
=> σ σ φ ∂
= − − ∇
∂t
A J
σ σ φ
µ
⎛ ⎞ ∂∇ × ∇ × + + ∇ =⎜ ⎟
∂⎝ ⎠
10
t
A A=>
Boundary and symmetry conditions
Vector potential on the outer surfaces of a solution region is
typically ∂= =∂
0 or 0 An
A A
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According to the first condition, the flux through the surface is
zero. The second condition states that the flux density isorthogonal to the surface.
= ± 21 A A
∂
A
Pole 2
Pole 1
− A
= A 0
= A 0
In the analysis of electricalmachines, symmetry of the field is
often utilised. The vector potential
is forced to be periodic on the
boundaries of the solution sector.
A
Ferromagnetic non-linearity of iron
Iron is used to increase the magnetic flux of a machine. Its
permeability is a very complicated function of flux density.=> The field equations become non-linear
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-1.50
-1.00
-0.50
0.00
0.50
1.00
1.50
-1500 -1000 -500 0 500 1000 1500
H [A/m]
B
[ T ]
Iron sample in alternating field Field in the yoke of electrical machine
Modelling motion
Usually, there is a slotting
on both stator and rotor surfaces, and there is no
reference frame in which
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the motion would not affect
the material properties.
The stator and rotor fieldsmust be solved in their own
reference frames and
forced to be continuous
over the air gap.
σ σ φ µ
⎛ ⎞ ∂∇ × ∇ × + + ∇ =⎜ ⎟ ∂⎝ ⎠
10
t
A A
σ σ φ µ
⎛ ⎞ ∂∇ × ∇ × + + ∇ =⎜ ⎟ ∂⎝ ⎠
1 ' ' ' 0' t
A' A'
In air gap: = A A'
Simplification I – Two-dimensional field
The solution of a time- and motion-dependent, non-linear,
three-dimensional magnetic field requires too muchcomputation power and time.
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The field is usually assumed to be two-dimensional,
independent from the coordinate parallel to shaft of themachine (z-direction).
The solution task is reduced significantly as the vector
potential has only one component and the scalar potential is asimple, linear function of the coordinate parallel to the shaft.
( )φ φ
⎧⎪ =
⎪ =⎨⎪⎪ = − +⎩
0
( , , )
( , , )z
z
A x y t
J x y t
u tz
l
A e
J e=>
( )σ σ
µ
∂⎡ ⎤−∇ ⋅ ∇ + =⎢ ⎥ ∂⎣ ⎦
( , , )1( , , )
u t A x y t A x y t
t l
Circuit equations
Electrical machines are typically supplied from voltage
sources. The voltage induced in a winding should beintegrated along the winding
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