Question 6.35: 300 kg/min of steam (2 bar, 0.98 dry) flows through a given ...
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)
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.
Blade height, h :
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.
