Question 6.13: The inlet and outlet angles of blades of a reaction turbine ......
The inlet and outlet angles of blades of a reaction turbine are 25 and 18°, respectively. The pressure and temperature of steam at the inlet to the turbine are 5 bar and 250°C. If the steam flow rate is 10 kg/s and the rotor diameter is 0.72 m, find the blade height and power developed. The velocity of steam at the exit from the fixed blades is 90 m/s.
Figure 6.30 shows the velocity triangles.
α_1 = β_2 = 18°, and α_2 = β_1 = 25°
C_1 = 90 m/s
From the velocity triangle,
C_{w1} = C_1 \cos(α_1) = 90 cos 18° = 85.6 m/s
C_{a1} = CD = C_1 \sinα_1 = 90 sin 18° = 27.8 m/s
From triangle BDC
BD = \frac{C_{a1}} {\sin (β_1)} = \frac{27.8}{\sin (25°)} = \frac{27.8} {0.423} = 65.72 m/s
Hence blade velocity is given by:
U = C_{w1} – BD = 85.6 – 65.62 = 19.98 m/s.
Applying the cosine rule,
V_{1}^2 = C^2_1 + U^2 – 2C_1U\cosα_1
= 90² + 19.98² – (2) × (90) × (19.98) cos18°
V_1 = 71.27 m/s
From triangle AEF,
C_{w2} = C_2 \cos(α_2) = 71.27 cos 25° = 64.59 m/s
Change in the velocity of whirl:
ΔC_w = C_{w1} + C_{w2} = 85.6 + 64.59 = 150.19 m/s
Power developed by the rotor:
P = \dot{m}UΔC_w = \frac{(10) × (19.98) × (150.19)} {1000} = 30 kW
From superheated steam tables at 5 bar, 250°C, the specific volume of steam is:
v = 0.4744 m³/kg
Blade height is given by the volume of flow equation:
v = πDhC_a
where C_a is the velocity of flow and h is the blade height. Therefore,
0.4744 = π × (0.72) × (h) × (27.8), and
h = 0.0075 m or 0.75 cm
