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218 Renewable Energy Devices and Systems with Simulations in MATLAB and ANSYS ®
®
b. The first equation in (9.10) may be solved for the rotor flux ψ r with V s and I s in phase as real
numbers:
+ 06
3200 j − 30.
)
ψ =− −(0 03413 + j0 4096 541 9 × × =−072. + + j606 Wb
.
.
.
.
r 3 314 3
Now, from the second equation of (9.10)
3 0 03413 ×314
.
×
.
V = 0 03413 × × 541 9 . − − j025 314
.
r
× 413
.
306 3063.
.
.
.
.
j
j
×− ( 072 + 6 06 ) =−456 8. − 6273
r
note that the real rotor voltage is V = a rs V = 4 V .
r
r
r
The rotor current I r is from (9.9):
.
341
−×
.
−072 + j6 06 3 × 541 9 .
.
I = 3413. 314 =−552 94. + j182 3 A
.
r
× 06
.
314
The rotor active power P r is simply
r (
*
.
.
.
P r = 3Re V I r ) = 3Re − ( 456 8 − j6273 ) ×− ( 552 94. − j182 3 ) = 723 45. kW
The rotor reactive power Q r (at the slip frequency) is
*
r
Q r = 3Im VI ( rr ) = 353 88. kVAR
The total power at max speed is P t = P r + P s = 3000 723 45 3723 45 kW = 372 MW.
=
+
.
.
.
2
The rotor-side converter has to be designed for S r = P r + Q r = 723 45 2 + 353 88 2 =
2
r
.
.
805 36 kVA = 080 MVA.
.
.
9.2.3 DFIG dq Model and Control
The dq model of a DFIG for generator operation in synchronous coordinates is [4]
dψ d =− V d − R i s d + ωψ dψ q =− V q − R i s q ωψ
−
dt 1 q ; dt 1 d
dψ dr =− V dr − R i rdr +( ω − r ω ψ ) dψ qr =− V qr − R i rqr −( ω − r ω ψ )
dt 1 qr ; dt 1 dr
rdrqr +
sd q +
m dr qr ;
ψ dq, = Li , Li , ψ dr qr, = Li , Li , (9.13)
m ds qs
3
T e = p (ψ i q − ψ i d)
2 1 d q
J dω
⋅ r = T e + T mech ; T e < 0 for generating
p 1 dt