Question 6.39: (Impulse reaction turbine). The following data relate to a s...

Steam velocity coming out of nozzle, C_{1} = 245 m/s
Nozzle angle,                                 α = 20°
Blade mean speed,                      C_{bl} = 145 m/s
Speed of the rotor,                       N = 3000 r.p.m.
Blade height,                                 h = 10 cm = 0.1 m
Specific volume of steam at nozzle outlet or blade inlet, v_{1} = 3.45 m³/kg
Specific volume of steam at blade outlet, v_{0} = 3.95 m³/kg
Power developed by the turbine = 287 kW
Efficiency of nozzle and blades combinedly = 90%
Carry over co-efficient, ψ = 0.82
Blade speed,                        C_{bl} =  \frac{πDN}{60}

∴                                                D =   \frac{ 60        C_{bl}  }{πN}  =  \frac{60    ×     145}{π    ×     3000}  = 0.923 m

Mass flow rate,                              \dot{m}_{s}  =  \frac{  C_{f_1}     ×    πDh    }{v_{1}}   =  \frac{C_{1}     sin  α    ×    πDh    }{v_{1}}

=  \frac{245   ×    sin    20°   ×    π        ×        0.923   ×    0.1}{3.45}   =  7.04  kg/s

Also                               \dot{m}_{s}  =  \frac{  C_{f_0}      πDh    }{v_{0}}

∴                                                    C_{f_0} =  \frac{\dot{m}_{s}    v_{0}}{πDh}    =  \frac{7.04       ×      3.95}{π       ×      0.923       ×      0.1} =  95.9 m/s

The power is given by,                      P = \frac{\dot{m}_{s}   ×    C_{bl}    ×    C_{w}}{1000}

287 =  \frac{7.04     ×   145    ×    C_{w} }{1000}

∴                                                              C_{w}  = \frac{287   ×    1000}{7.04     ×   145    }   = 281  m/s

Now draw velocity triangles as follows :
Select a suitable scale (say 1 cm = 25 m/s)
•  Draw LM = blade velocity = 145 m/s ; ∠MLS = nozzle angle = 20° Join MS to complete the inlet ∆LMS.
•  Draw a perpendicular from S which cuts the line through LM at point P. Mark the point Q such that PQ =   C_{w}   = 281 m/s.
•  Draw a perpendicular through point Q and the point N as QN = 95.9 m/s. Join LN and MN to complete the outlet velocity triangle.

From the velocity triangles ;

(i) Heat drop in each stage, (∆h)_{stage} :
Heat drop in fixed blades ( ∆h_{f} )

=  \frac{ C_{1}²   –      ψ    C_{0}²}{2  ×    η_{nozzle}}   , where ψ is a carry over co-efficient

=  \frac{(245)²     –   0.82    ×     (105)²}{2   ×     0.9    ×    1000}   =   28.32  kJ/kg

Heat drop in moving blades ( ∆h_{m} )

=  \frac{ C_{r_0}²   –        C_{r_1}²}{2  ×    η_{nozzle}}   =  \frac{(217.5)²     –    (117.5)²}{2   ×     0.9    ×    1000} =    18.61   kJ/kg

Total heat drop in a stage,

( ∆h)_{stage}  =     ∆h_{f}   +   ∆h_{m}   = 28.32 + 18.61 = 46.93 kJ/kg.

(ii) Degree of reaction, R_{d} :

R_{d}  =  \frac{ ∆h_{m} }{ ∆h_{m}   +    ∆h_{f} }  =  \frac{18.61 }{18.61  +   28.32}    = 0.396. 

(iii) Stage efficiency,  η_{stage} :

Work done per kg of steam

= \frac{C_{bl}    ×   C_{w}}{1000}  =  \frac{145    ×     281}{1000}    = 40.74 kJ/kg of steam

∴                                                    η_{stage}   =  \frac{Work  done   per    kg    of    steam}{Total  heat   drop    in  a   stage}    = \frac{40.74}{46.93} =  0.868   or   86.8%.