Question 6.DE.9: In a Parson’s turbine, the axial velocity of flow of steam i......
In a Parson’s turbine, the axial velocity of flow of steam is 0.5 times the mean blade speed. The outlet angle of the blade is 20°, diameter of the ring is 1.30 m and the rotational speed is 3000 rpm. Determine the inlet angles of the blades and power developed if dry saturated steam at 0.5 MPa passes through the blades where blade height is 6 cm. Neglect the effect of the blade thickness.
The blade speed, U = \frac{\pi DN} {60} = \frac{\pi × (1.30) × (3000)} {60} = 204 m/s
Velocity of flow, C_a = (0.5) × (204) = 102 m/s
Draw lines AB and CD parallel to each other Fig. 6.26 at the distance of
102 m/s, i.e., velocity of flow, C_{a1} = 102 m/s.
At any point B, construct an angle α_2 = 20° to intersect line CD at point C. Thus, the velocity triangle at the outlet is completed. For Parson’s turbine,
α_1 = β_2, β_1 = α_2, C_1 = V_2, and V_1 = C_2.
By measurement,
ΔC_w = C_{w1} + C_{w2} = 280.26 + 76.23 = 356.5 m/s
The inlet angles are 53.22°.Specific volume of vapor at 0.5 MPa, from the
steam tables, is
v_g = 0.3749 m³/kg
Therefore the mass flow is given by:
\dot{m} = \frac{AC_2} {x_2v_{g2}} = \frac{\pi × (1.30) × (6) × (102)}{(100) × (0.3749)} = 66.7 kg/s
Power developed:
P = \frac{\dot{m}UΔC_w} {1000} = \frac{(66.7) × (356.5) × (102)} {1000} = 2425.4 kW
