Page 203 - Fluid Mechanics and Thermodynamics of Turbomachinery
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184 Fluid Mechanics, Thermodynamics of Turbomachinery
                            In a particular problem the quantities c x2 , c  2 , ˇ 3 are known functions of radius
                          and  can be specified. Equation (6.34) is thus a first order differential equation in
                          which c x3 is unknown and may best be solved, in the general case, by numerical
                          iteration. This procedure requires a guessed value of c x3 at the hub and, by applying
                          eqn. (6.34) to a small interval of radius r, a new value of c x3 at radius r h C r is
                          found. By repeating this calculation for successive increments of radius a complete
                          velocity profile c x3 can be determined. Using the continuity relation
                              Z          Z
                                r t        r t
                                 c x3 rdr D  c x2 rdr,
                               r h         r h
                          this initial velocity distribution can be integrated and a new, more accurate, estimate
                          of c x3 at the hub then found. Using this value of c x3 the step-by-step procedure is
                          repeated as described and again checked by continuity. This iterative process is
                          normally rapidly convergent and, in most cases, three cycles of the calculation
                          enables a sufficiently accurate exit velocity profile to be found.
                            The off-design performance may be obtained by making the approximation that
                          the rotor relative exit angle ˇ 3 and the nozzle exit angle ˛ 2 remain constant at a
                          particular radius with a change in mass flow. This approximation is not unrealistic
                          as cascade data (see Chapter 3) suggest that fluid angles at outlet from a blade row
                          alter very little with change in incidence up to the stall point.
                            Although any type of flow through a stage may be successfully treated using this
                          method, rather more elegant solutions in closed form can be obtained for a few
                          special cases. One such case is outlined below for a free-vortex turbine stage whilst
                          other cases are already covered by eqns. (6.21) (6.23).


                          Free-vortex turbine stage

                            Suppose, for simplicity, a free-vortex stage is considered where, at the design
                          point, the flow at rotor exit is completely axial (i.e. without swirl). At stage entry
                          the flow is again supposed completely axial and of constant stagnation enthalpy h 01 .
                          Free-vortex conditions prevail at entry to the rotor, rc  2 D rc x2 tan a 2 D constant.
                          The problem is to find how the axial velocity distribution at rotor exit varies as the
                          mass flow is altered away from the design value.
                            At off-design conditions the relative rotor exit angle ˇ 3 is assumed to remain
                                                                  Ł
                                           Ł
                          equal to the value ˇ at the design mass flow ( denotes design conditions). Thus,
                          referring to the velocity triangles in Figure 6.7, at off-design conditions the swirl
                          velocity c  3 is evidently non-zero,
                              c  3 D c x3 tan ˇ 3  U
                                 D c x3 tan ˇ Ł 3  r.                                    .6.35/
                          At the design condition, c Ł  D 0 and so
                                                 3
                                     Ł
                               Ł
                              c tan ˇ D r.                                               (6.36)
                               x3    3
                          Combining eqns. (6.35) and (6.36)

                                        c x3
                              c  3 D r      1 .                                          (6.37)
                                        c Ł
                                         x3
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