Page 128 - Fluid Mechanics and Thermodynamics of Turbomachinery
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Axial-flow Turbines: Two-dimensional Theory  109
                                        3.0
                                                 h tt  =  0.82  120°  100° 0.86  0.88
                                                          0.84
                                                    e =  140°
                                       Stage loading coefficient, y =  DW/U 2  2.0  0.92  0.90  60° 40°

                                                                     80°






                                        1.0





                                                              0.86  0.84  0.82  20°

                                                       0.5          1.0          1.5
                                                      Flow coefficient, f = c /U
                                                                     x
                          FIG. 4.10. Design point total-to-total efficiency and deflection angle contours for a turbine
                                                 stage of 50 per cent reaction.

                            Referring to Figure 4.5, for zero reaction ˇ 2 D ˇ 3 , and from eqn. (4.21)

                                                                    1/ .
                              tan ˛ 2 D 1/  C tan ˇ 2 and tan ˛ 3 D tan ˇ 3
                                         2
                          Also,   D W/U D  .tan ˛ 2 C tan ˛ 3 / D  .tan ˇ 2 C tan ˇ 3 / D 2  tan ˇ 2 ,

                              ∴ tan ˇ 2 D  .
                                        2

                          Thus, using the above expressions:
                              tan ˛ 2 D . /2 C 1//  and tan ˛ 3 D . /2  1// .
                          From these expressions the flow angles can be calculated if values for   and   are
                          specified. From an inspection of the velocity diagram,
                                                  2
                                                                      2
                                                              2
                                                                                      2
                                                                                   2
                                                      2
                              c 2 D c x sec ˛ 2 , hence c D c .1 C tan ˛ 2 / D c [1 C . /2 C 1/ /  ],
                                                  2   x               x
                                                                      2
                                                       2
                                                              2
                                                                                 2
                                                  2
                              w 3 D c x sec ˇ 3 , hence w D c .1 C tan ˇ 3 / D c [1 C . /2 / ].
                                                  3    x              x
                          Substituting the above expressions into eqn. (4.9a):
                                          2
                               1         R w C   N c 2 2
                                          3
                                 D 1 C        2   ,
                                tt        2 U
                                          (   "          #      "              #)
                                                         2                    2
                               1        1       2                 2
                                 D 1 C        R   C        C   N   C 1 C          .
                                tt     2              2                    2
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