6 folier pr. side - NTNU

Trondheim 2000
NTNU
Slide 253
Transformasjon for 3-fase viklinger
■ For trefase viklingen finner man helt tilsvarende:
[ ] [ ]
[ ] T
k
s
k
s
k
s
k
s
k
k
s
k
k
s
k
k
s
s
T
sc
sb
sa
S
s
s
sc
s
sb
s
sa
s
I
I
I
I
I
I
I
I
I
I
I
I
c
I
b
I
a
I
3
2
I
γ
β
α
γ
β
α
=
γ
⋅
+
β
⋅
+
α
⋅
=
=
⋅
+
⋅
+
⋅
⋅
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
γ
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
θ
θ
−
=
β
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
θ
θ
=
α
1
0
0
0
cos
sin
0
sin
cos
k
k
k
k
k
k
k
[ ]
[ ]
[ ]
S
S
S
k
S
0
k
S
0
k
S
k
k
S
0
k
S
0
k
S
k
k
c
b
a
3
1
c
)
240
sin(
b
)
120
sin(
a
sin
3
2
c
)
240
cos(
b
)
120
cos(
a
cos
3
2
+
+
=
γ
⋅
−
θ
+
⋅
−
θ
+
⋅
θ
−
=
β
⋅
−
θ
+
⋅
−
θ
+
⋅
θ
=
α
Trondheim 2000
NTNU
Slide 254
Transformasjon for 3-fase viklinger
■ Finner da Park-transformasjonen:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
θ
−
−
θ
−
θ
−
−
θ
−
θ
θ
⋅
=
2
/
1
2
/
1
2
/
1
)
240
sin(
)
120
sin(
sin
)
240
cos(
)
120
cos(
cos
3
2 0
k
0
k
k
0
k
0
k
k
k
ss
�
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
θ
−
−
θ
−
θ
−
−
θ
θ
−
θ
=
−
1
)
240
sin(
)
240
cos(
1
)
120
sin(
)
120
cos(
1
sin
cos
0
k
0
k
0
k
0
k
k
k
k
ss
�
■ Den inverse Park-transformasjonen:
Trondheim 2000
NTNU
Slide 255
Transformasjon for 3-fase rotor viklinger
i en asynkronmaskin
■ Finner da Park-transformasjonen:
■ Den inverse Park-transformasjonen:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
θ
−
−
θ
−
θ
−
−
θ
−
θ
θ
⋅
=
2
/
1
2
/
1
2
/
1
)
240
sin(
)
120
sin(
sin
)
240
cos(
)
120
cos(
cos
3
2 0
r
0
r
r
0
r
0
r
r
k
rr
�
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
θ
−
−
θ
−
θ
−
−
θ
θ
−
θ
=
−
1
)
240
sin(
)
240
cos(
1
)
120
sin(
)
120
cos(
1
sin
cos
0
r
0
r
0
r
0
r
r
r
k
rr
�
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
= −
−
−
k
rr
k
ss
k
k
rr
k
ss
k
�
�
�
�
�
�
�
�
�
�
Trondheim 2000
NTNU
Slide 256
Transformasjon for rotor viklinger
i en synkronmaskin
■ Finner da transformasjonen:
■ Den inverse transformasjonen:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
=
1
0
0
0
1
0
0
0
1
r
rr
�
�
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
=
−
1
0
0
0
1
0
0
0
1
r
rr
�
�
r
ss
r
r
ss
r
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
=
−
−
�
�
�
�
�
�
�
�
�
�
Trondheim 2000
NTNU
Slide 257
Elektriske likninger og momentbalanse i
den transformerte modell
■ Fysikalsk modell:
■ Utledning av transformerte modell:
I
dt
d
I
U
SR
SR
SR
SR
SR
SR
SR
�
� =
Ψ
Ψ
+
= ( ) SR
SR
T
SR
e
I
I
2
p
M ⋅
θ
∂
∂
⋅
⋅
=
�
I
dt
d
I
U
U
SR
SR
r
SR
r
r
SR
r
SR
SR
r
SR
r
r
�
�
�
�
�
�
� =
Ψ
=
Ψ
Ψ
+
=
=
I
I
I
I
r
r
SR
SR
r
r
r
r
SR
SR
r
r
Ψ
=
Ψ
Ψ
=
Ψ
=
=
−
− �
�
�
�
( )
r
r
r
r
r
r
r
r
SR
r
r
r
r
r
r
r
r
r
r
SR
r
r
r
r
r
r
r
SR
r
SR
r
r
d
d
dt
d
I
U
dt
d
dt
d
I
U
dt
d
I
U
U
Ψ
θ
⋅
ω
+
Ψ
+
=
Ψ
+
Ψ
+
=
Ψ
+
=
=
−
−
−
−
−
−
−
−
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
θ
=
=
Ψ
⋅
ω
+
Ψ
+
=
−
−
d
d
dt
d
I
U
r
r
r
SR
r
r
r
r
r
r
r
�
�
�
�
�
�
�
�
�
Trondheim 2000
NTNU
Slide 258
Elektriske likninger og momentbalanse i
den transformerte modell……….
■ Transformerte modell:
■ Motstander og induktansmatrise:
θ
=
=
Ψ
⋅
ω
+
Ψ
+
=
−
−
d
d
dt
d
I
U
r
r
r
SR
r
r
r
r
r
r
r
�
�
�
�
�
�
�
�
�
I
I
r
SR
r
r
r
r
r
r
SR
r
SR
r
r −
−
=
=
=
Ψ
=
Ψ �
�
�
�
�
�
�
�
�
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⋅
⋅
⋅
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
Q
aQ
D
fD
aD
fD
f
af
0
aQ
q
aD
af
d
r
Q
D
f
s
s
s
r
L
0
0
0
L
2
/
3
0
0
L
L
0
0
L
2
/
3
0
L
L
0
0
L
2
/
3
0
0
0
L
0
0
L
0
0
0
L
0
0
L
L
0
0
L
R
0
0
0
0
0
0
R
0
0
0
0
0
0
R
0
0
0
0
0
0
R
0
0
0
0
0
0
R
0
0
0
0
0
0
R
�
�