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Study of control strategies of power electronics during faults in microgrids    127


                        k  1− k          k  1− k 
                I qL  =  Q *   v q +  +  v −  q   I ,  qS  =  Q *  v q +  −  v −  q   (7.42)                        IqL=Q*kqv⊥++1−kqv⊥−,   IqS=Q*kqv
                                
                                         
                      
                                                   
                                                 ⊥
                                            ⊥
                         ⊥
                              ⊥
                                                                                                                        ⊥+−1−kqv⊥−
                                 +
                    φ − φ −     φ −  φ −  π     φ −  φ −  π
                                                 +
                     +
               γ =         , γ =        +  , γ =       −                   (7.43)                                       γa=φ+−φ−2,   γb=φ+−φ−2+π3,   γc=
                 a
                       2     b     2     3   c     2     3                                                              φ+−φ−2−π3
           where Î m  represents the peak current in each phase; I pL  and I pS  are the values of the
           long and short axes of the active current ellipse; I qL  and I qS  are the values of the long
                                                 +
           and short axes of the reactive current ellipse; φ  and φ  are the phase angles of posi-
                                                       −
           tive- and negative-sequence voltage respectively. Take phase-A as an example, by sub-
           stituting Eqs. (7.41)–(7.43) into Eq. (7.40), the relationship among the peak current of
           phase-A Î a  and the power references can be expressed as:
                                                    q)
                         k v −  2  +  1 ( − k ) 2  v +  2  −  k 2 ( 1−  k cos(2 ) vγ  +  ⋅  v −  
                          2
                  0 = Q   q        q         q            a       
                      *2
                                         v +  2 ⋅  v −  2         
                                                                  
                                           pq)
                                k 2 (  +  k 2  −  4 kk sin(2 ) vγ  +  ⋅  v 
                                                           −
                         − PQ    p    q            a       
                             *
                           *
                                        v +  2 ⋅  v −  2   
                                                 p)
                      k v −  2 +  1 ( − k ) 2  v +  2 +  k 2 ( 1− k cos(2 ) vγ  +  ⋅  v −  
                       2
                 + P   p         p         p           a         −  Î  2  (7.44)                                      0 = Q * 2 k q 2 v − + 1 − k q v + − 2 k q 1 − k q
                                                                                                                                                  2
                                                                                                                                      2
                                                                                                                                              2
                   *2
                                      v +  2 ⋅  v −  2            a                                                                        2     2
                                                                                                                      cos (2 γ a )v+ ⋅ v − v+ ⋅ v − − P * Q *2
                                                                                                                                                        2
                                                                                                                        kp+2kq−4kpkqsin(2γa)v+⋅v−v+ ⋅v−-
                                                                                                                                               2
                                                                                                                                                   2
                                                                                                                        2  + P *2 k p 2v − +1 − k p v+ +2 k p 1 − k
                                                                                                                                      2
                               ∗
              Given the values of Q , k  and k , the expression in Eq. (7.44) can be regarded as a                      pcos(2γa)v+⋅v−v+ ⋅v− −Îa2
                                                                                                                                       2
                                                                                                                                           2
                                       q
                                  p
                               ∗
           quadratic equation with P  as an unknown variable. The solution of this quadratic equa-
           tion gives the maximum allowed active power P a lim  to comply with the current limit of
           phase-A. With γ  changed to γ  and γ , the maximum allowed active power to guarantee
                        a
                                        c
                                   b
           a safe current in phase-B and phase-C, namely P b lim  and P c lim , can be obtained using the
           same procedure. Finally, the limit on active power under fault conditions is restricted by:
                          lim
                              lim
                                  lim
                P max  = min{ P , P , P }                                  (7.45)                                       Pmax=minPalim,Pblim,Pclim
                                  c
                              b
                          a
              To illustrate the effectiveness of this current limitation method, the short-circuit
           response of a VSI to a phase A-B fault with 0.1 Ω fault resistance is plotted in Fig. 7.6
                                              ∗
           for the same scenarios of Fig. 7.5. With Q  set to zero, the current in each phase is
           restricted within 61.23 A.
           3.4.2  Flexible oscillating power control
           The current limitation method presented above is only valid for flexible positive- and
           negative-sequence power control because current references Eqs. (7.28) and (7.29)
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