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## 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.)

## 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

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.