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## Q. 6.35

300 kg/min of steam (2 bar, 0.98 dry) flows through a given stage of a reaction turbine. The exit angle of fixed blades as well as moving blades is 20° and 3.68 kW of power is developed. If the rotor speed is 360 r.p.m. and tip leakage is 5 per cent, calculate the mean drum diameter and the blade height. The axial flow velocity is 0.8 times the blade velocity.                                                                                                                                                                     (Roorkee University)

## Verified Solution

Rate of flow of steam through the turbine,  $\dot{m}_{s} = \frac{300}{60}$  =  5  kg/s
Pressure and condition of steam, p = 2 bar, x = 0.98.
The exit angles of fixed blades as well as moving blades, α = φ = 20°
Power developed,                             P = 3.68 kW
Speed of the rotor,                            N = 360 r.p.m.
Tip leakage                                          = 5 per cent
Axial flow velocity,                              $C_{f} = 0.8 C_{bl}$  (blade velocity)
Refer Fig. 49.

Mean drum diameter, D :

Mean blade velocity,                                $C_{bl} = \frac{πDN}{60} = \frac{πD × 360}{60}$  = 18.85 D m/s

Power developed,                             P =  $\frac{\dot{m}_{s} C_{bl} C_{w}}{1000}$

or                                            3.67 = $\frac{(5 × 0.95) × 18.85 D × C_{w} }{1000}$

∴                                                            $C_{w} = \frac{3.67 × 1000}{(5 × 0.95) × 18.85 D} = \frac{40.988}{D}$

Assuming Parson’s reaction turbine, we have

$C_{f_1} = C_{1} sin α or C_{1} = \frac{C_{f_1} }{sin α} = \frac{0.8 C_{bl} }{sin α} = \frac{0.8 × 18.85 D}{sin 20°}$ = 44.091 D
$(C_{f_1} = C_{f_0} = C_{f})$

Also,                                                $C_{w} = 2 C_{1} cos α – C_{bl} or \frac{40.988}{D} = 2 × 44.091 D \cos 20° – 18.85 D = 64.01 D$

or                         40.988 = 64.01 D²    or    D = 0.8 m or 800 mm.

Mean steam flow rate,    $\dot{m}_{s} = \frac{πDhC_{f}}{x v_{g}}$
or                            (5 × 0.95) =  $\frac{π × 0.8 × h × C_{1} sin α}{0.98 × 0.885} = \frac{π × 0.8 × h × (44.091 D × sin 20°)}{0.98 × 0.885}$
(At 2 bar : $v_{g}$ = 0.885 kg/m³)
or                                h =  $\frac{(5 × 0.95) × 0.98 × 0.885}{π × 0.8 × (44.091 × 0.80 × 0.3420)}$   = 0.1359 m        or        135.9 mm. 