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Chapter 6

Q. 6.34

A 50% reaction turbine (with symmetrical velocity triangles) running at 400 r.p.m. has the exit angle of the blades as 20° and the velocity of steam relative to the blades at the exit is 1.35 times the mean blade speed. The steam flow rate is 8.33 kg/s and at a particular stage the specific volume is 1.381 m³/kg. Calculate for this stage :
(i) A suitable blade height, assuming the rotor mean diameter 12 times the blade height, and
(ii) The diagram work.                                                                                 (N.U.)

Step-by-Step

Verified Solution

Speed,                                     N = 400 r.p.m. ; α = 20°
C_{r_0}    =   C_{1}    =    1.35   C_{bl}  ;    \dot{m}_{s}   = 8.33 kg/s
v = 1.381 m³/kg ; D = 12 h

(i) Blade height, h :
Refer Fig. 49.

Axial flow velocity,                       C_{f_1}  =    C_{f_0}   =   C_{f}   =   C_{1} \sin  α

= 1.35  C_{bl}  \sin  20°
= 0.4617   C_{bl}

Area of flow,                  A = πDh = πD ×  \frac{D}{12}  =  \frac{πD²}{12}

Mass flow rate,                                                    \dot{m}  =    \frac{A C_{f}}{v}    or           8.33 =  \frac{A  ×    0.4617   C_{bl}}{1.381}

or                                                \frac{8.33   ×    1.381 }{0.4917}       =  A   ×    C_{bl}   =  \frac{πD²}{12}     ×    \frac{πDN}{60}

or                                              24.916 =  \frac{π²D³   ×    400}{720}                                                     or                   D³  = \frac{24.919      ×     720}{π²     ×    400  }                  or     D = 1.656   m

∴               Blade height,                 h =  \frac{D}{12}  =  \frac{1.656}{12}     =  0.138  m        or          138  mm.

(ii) The diagram work :

Diagram work                             =  \dot{m}     ×      C_{bl}    (C_{w_1}    +    C_{w_0})

\dot{m}     ×      C_{bl}    (2  C_{1}    \cos   α    –  C_{bl})

= 8.33     ×      C_{bl}  (2    ×     1.35    C_{bl}  \cos  20°    –    C_{bl})   

8.33     ×      C_{bl}²  (2    ×     1.35  \cos  20°    –   1  )

= 8.33 ×  ( \frac{πDN}{60} )²   ×    1.537   =  8.33 ×   ( \frac{π   ×      1.656   ×    400}{60} )²  ×    1.537

= 15401.2 W       or      15.4 kW.

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