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Tidal Energy Chapter | 3 75


                                      +
                                  u dis , p dis
                                                                       u w , p o
                    u o , p o
                                                       u h
                                                    u t







                                             −
                                         u dis , p dis





                      Upstream, S1    Turbine (disk), S2        Wake, S3
             FIG. 3.27  Stream tube around a horizontal axis tidal turbine. The cross-sections for implementa-
             tion of the momentum and energy equations have been also shown (S1, S2, and S3).



             unknown, we can apply the energy equation in the upstream (between S1 and
             S2) or the downstream part (between S2 and S3) of the disk.
                Applying the energy equation between S1 and S2, leads to
                        u 2 o  p o  u 2 dis  p +  p o  u 2 dis  p +  u 2 o
                                         dis
                                                           dis
                           +    =     +     ⇒    =     +     −         (3.37)
                        2g   ρg    2g   ρg    ρg    2g    ρg   2g
             Similarly, for the downstream sections (S2 and S3), we have
                        u 2  p −   u 2   p o   p o  u 2   p −  u 2
                         dis  dis   w                dis   dis  w
                           +     =    +     ⇒    =     +     −         (3.38)
                        2g    ρg    2g   ρg   ρg    2g    ρg   2g
             therefore
                             u 2   p +   u 2  u 2   p −   u 2
                              dis   dis   o    dis   dis   w
                                 +     −    =     +    −               (3.39)
                              2g   ρg    2g   2g    ρg    2g
             which leads to
                                             1   2   2
                                  p +  − p −  =  ρ(u − u )             (3.40)
                                   dis  dis      o   w
                                             2
                By equating the RHS of Eqs (3.36), (3.40), we can write
                                1   2    2
                                 ρ(u − u ) = ρu dis (u o − u w )       (3.41)
                                    o
                                         w
                                2
             or
                                1
                                 (u o + u w ) = u dis = u o (1 − a)    (3.42)
                                2
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