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224             Renewable Energy Devices and Systems with Simulations in MATLAB  and ANSYS ®
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                                           jω ψ   R i      P <0
                                                           s
                                                   s s
                                            1 s
                                                           Q >0
                                                            s
                                                           S<0
                                                 V s
                                               i d  L i
                                                    d d
                                                       jL i
                                                        q q
                                         ji q        ψ
                                                 i s  s
            FIGURE 9.13  CRIG space phasor diagram for generator mode (steady state).


              With ψ s =  Li  jL scq i  and neglecting the stator copper losses,
                        s d +
                                  3      *   3        s ( )  3
                                                       *
                                      VI s =
                              P s ≈  Re ( )    ω 1 Im ψ  I s =  1 ω  ( L s − L sc) i i q
                                        s
                                                                       d
                                  2          2             2                          (9.21)
                                                              (
                                  3       *   3         *
                                       VI s =
                              Q s =  Im ( ) )   1 ω Re ( )  = 3ω L i 2  + L i 2 )
                                                     ψ I
                                                        s
                                        s
                                  2           2       s      1  sd   sc q
              Both ideal powers may be controlled by the variation i  > 0 and i  < 0 (generating), but as dem-
                                                          d
                                                                  q
            onstrated in FOC of the cage rotor for IMs, i  and i  may be varied separately (decoupled). So P  is
                                                d
                                                                                         s
                                                     q
            controlled by i  control and Q  by i  control. Besides FOC, direct torque (power) and flux control
                        q
                                    s
                                        d
            may be applied to CRIG as done earlier for IM drives [16].
              Example 9.2
              Let us consider a large power CRIG with the following data V n  = 3200 V (Y), S n  = 3 MVA, r s  = 0.015 pu,
              r r  = 0.0125 pu, l sl  = l rl  = 0.05 pu, and l m  = 3 pu, with six poles generating at 50 Hz with a slip S = −0.015.
              Calculate the resistances and reactances in Ω, stator flux, i d  and i q , and electromagnetic torque and
              powers.
              Solution
              Approximately ψ s ≈  s V  =  3200 2 /  3  = 8 306 Wb.  .
                                     π
                             ω 1    250
                                       3       3200 2
              The nominal reactance X n =  V l n  =  V l n  =  6  = .341Ω.
                                                ×
                                     I n   S n  310
                                                                                  Ω
                                                             ×
                                                 Ω
                                                                  =
                                                                       Ω
                                      Ω
                                                                .
                   R s = 0 015 ×  X n = 0 0512 ,  R r = 0 0427 ,  X sl =  X rl = 05 3410 17 ,  X m =10024 ,
                                                                    .
                        .
                                            .
                                 .
                                                                               .
                                                           .
                   L m =  0 0326 , L s =H  L r =  0 033 , L sc =H  1 087 mH
                                                 .
                        .
                                       .
                                      s d) +(
              but in rotor coordinates ψ s = ( Li  2  Li  2  Sω 1  L r  i d .
                                             scq)  and i q =−
                                                           R r
                 From the previous two equations, we may calculate i d  and i q  as
                                                      .
                                                                            .
                 i d =      s ψ    =                 8 306                =  8 306  =  250 A
                                                                            .
                              L r  2               −3          0 032  2  0 033
                                                                   .
                                                       2
                      2
                     L s +  L sc   sω 1    0 033 2  +( 1 087.  ×10  )  −0015 314  
                          2
                                                              ×
                                                             5
                                                          .
                                       .
                                                                   .
                              R r                              0 0427 
                                         .
                                  ×
              Now i q =−  Sω 1  Li rd  = 0 015 314  ×  0 033  × 250 A  = 910 86 A.
                               .
                                                        .
                                         .
                          R r           0 0427
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