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Three-dimensional Flows in Axial Turbomachines 179
                          From the radial equilibrium solution in eqns. (6.19), after some rearrangement,
                              c 2  c 2              2a
                                                            r
                               x1   x2  D  A 1  A 2  C  ln     ,
                                4b        2  t            r t
                          where
                                   W     C p T 0
                                t D    D        .
                                            2 2
                                   U 2      r
                                     t        t
                                                       1
                          In the above example, 1  a/ D ,  t D 0.18
                                                       2
                               0
                              R D 0.778 C ln.r/r t /.
                          The true reaction variation is shown in Figure 6.5 and it is evident that eqn. (6.17)
                          is invalid as a result of axial velocity changes.


                          The direct problem

                            The flow angle variation is specified in the direct problem and the radial equi-
                          librium equation enables the solution of c x and c   to be found. The general radial
                          equilibrium equation can be written in the form

                                     ds   c 2   dc
                              dh 0
                                    T   D    C c
                               dr    dr    r    dr
                                               2
                                           2
                                          c sin ˛    dc
                                        D         C c  ,                                  .6.20/
                                             r       dr
                          as c   D c sin ˛.
                            If both dh 0 /dr and ds/dr are zero, eqn. (6.20) integrated gives
                                      Z
                                              dr
                                           2
                              log c D    sin ˛  C constant
                                              r
                          or, if c D c m at r D r m , then
                                        Z   r
                               c               2  dr
                                 D exp       sin ˛    .                                   (6.21)
                              c m                  r
                                           r m
                          If the flow angle ˛ is held constant, eqn. (6.21) simplifies still further,
                                                       2
                                                      sin ˛
                               c    c x   c      r
                                 D     D     D                                            (6.22)
                              c m  c xm  c  m    r m
                          The vortex distribution represented by eqn. (6.22) is frequently employed in practice
                          as untwisted blades are relatively simple to manufacture.
                            The general solution of eqn. (6.20) can be found by introducing a suitable inte-
                                                                                  R    2
                          grating factor into the equation. Multiplying throughout by exp[2  sin ˛dr/r]it
                          follows that
                                          Z                                   Z
                               d   2           2             dh 0   ds             2
                                  c exp 2   sin ˛dr/r   D 2        T    exp 2    sin ˛dr/r .
                              dr                              dr    dr
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