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218             Renewable Energy Devices and Systems with Simulations in MATLAB  and ANSYS ®
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                 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
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