Question 6.38: The following data relate to a stage of reaction turbine : M...

Mean rotor diameter, D = 1.5 m
Speed ratio,                   ρ =  \frac{C_{bl}}{C_{1}} = 0.72
Blade outlet angle         = 20°
Rotor speed,                  N = 3000 r.p.m.
(This example solved purely by calculations (Fig. 52) is not drawn to scale.)
(i) Diagram efficiency :
Blade velocity,                                C_{bl} =  \frac{πDN}{60}  =  \frac{π      ×    1.5     ×    3000}{60}  = 235.6 m/s
Speed ratio,                       ρ =  \frac{C_{bl}}{C_{1}} = 0.72

∴                                                        C_{1}  =  \frac{C_{bl}}{0.72}  =  \frac{235.6}{0.72}    = 327.2 m/s
Assuming that velocity triangles are symmetrical
α = φ = 20°
From the velocity ∆LMS

= 100     \sqrt{ 10.7    +   5.55    –    14.48 }   = 133 m/s

i.e.,                                                      C_{r_1}    = 133 m/s

Work done per kg of steam                    =  C_{bl}   C_{w}   =    C_{bl}    (2  C_{1}   \ cos    α     –      C_{bl})

= 235.6 (2 × 327.2 \cos  20° – 235.6) = 89371.3 Nm.

Energy supplied per kg of steam

=  \frac{C_{1}²     +    C_{r_0}²     –      C_{r_1}²}{2}

= \frac{2  C_{1}²    –      C_{r_1}²}{2}                                                                                                 ( ∵  C_{1}   =  C_{r_0})

=  \frac{2  ×     (327.2)²   –     (133)² }{2}    =   98215.3   Nm

∴       Diagram efficiency      =  \frac{89371.3}{ 98215.3 }   = 0.91 = 91%.

(ii) Percentage increase in diagram efficiency :

For the best diagram efficiency (maximum), the required condition is

ρ =  \frac{C_{bl}}{C_{1}} = \cos    α

∴                                              C_{bl}  =  C_{1}   \cos  α = 372.2  \cos  20° = 307.46  m/s 

For this blade speed, the value of C_{r_1}  is again calculated by using eqn. (i),

= 100  \sqrt{10.7   +    9.45    –     18.906}  = 111.5 m/s

Diagram efficiency                                = \frac{2    C_{bl}   (2  C_{1}    \cos    α    –  C_{bl}  )}{(C_{1}²   +    C_{r_0}²     –     C_{r_1}²  }

= \frac{2   ×      307.46  (2   ×      327.2      \cos    20°     –    307.46)}{(327.2)²   +     (327.2)²    –    (111.5)²  }   =  0.937 or 93.7%

Percentage increase in diagram efficiency

=  \frac{0.937    –     0.91}{0.91}   = 0.0296     or     2.96%.

The diagram efficiency for the speed can also becalculated by using relation

The best theoretical speed of the rotor is given by,

C_{bl} =  \frac{πDN}{60}          or               N =    \frac{ 60        C_{bl}  }{πD}  =  \frac{60    ×     307.46}{π    ×     1.5}  =  3914.7 r.p.m.