Question 6.DE.14: From the following data, for 50% reaction steam turbine, det......
From the following data, for 50% reaction steam turbine, determine the blade height:
RPM: 440
Power developed: 5.5 MW
Steam mass flow rate: 6.8 kg/kW – h
Stage absolute pressure: 0.90 bar
Steam dryness fraction: 0.95
Exit angles of the blades: 70°
(angle measured from the axial flow direction).
The outlet relative velocity of steam is 1.2 times the mean blade speed. The ratio of the rotor hub diameter to blade height is 14.5.
Figure 6.31 shows the velocity triangles.
From the velocity diagram,
V_2 = 1.2 U
C_{a2} = V_2 \cos(β_2)
= 1.2U cos 70°
= 0.41U m/s
At mean diameter,
U = \frac{\pi DN} {60} = \frac{2\pi N(Dh + h)}{(60) \times (2)}
where D_h is the rotor diameter at the hub and h is the blade height.
Substituting the value of U in the above equation,
C_{a2} = \frac{(0.41) \times (2\pi) \times (440)(14.5h + h)} {(2) \times (60)} = 146.45 h m/s
Annular area of flow is given by:
A = \pi h(D_h + h) = \pi h(14.5h + h)
or
A = 15.5πh²
Specific volume of saturated steam at 0.90 bar, v_g = 1.869 m³/kg.
Then the specific volume of steam = (1.869) × (0.95) = 1.776 m³/kg.
The mass flow rate is given by:
\dot{m} = \frac{(5.5) × (10^3) × (6.8)} {3600} = 10.39 kg/s
But,
\dot{m} = \frac{C_{a2}A} {v} = \frac{C_{a2}15.5\pi h^2} {v}
Therefore:
10.39 = \frac{(146.45) × (h) × (15.5) × (\pi h^2)} {1.776}
or:
h³ = 0.00259, and h = 0.137 m
