Chapter 6
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)
Step-by-Step
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.
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.
