Question 6.DE.15: From the following data for a two-row velocity compounded im......
From the following data for a two-row velocity compounded impulse turbine, determine the power developed and the diagram efficiency:
Blade speed: 115 m/s
Velocity of steam exiting the nozzle: 590 m/s
Nozzle efflux angle: 18°
Outlet angle from first moving blades: 37°
Blade velocity coefficient (all blades): 0.9
Figure 6.32 shows the velocity triangles.
Graphical solution:
U = 115 m/s
C_1 = 590 m/s
α_1 = 18°
β_2 = 20°
The velocity diagrams are drawn to scale, as shown in Fig. 6.33, and the relative velocity:
V_1 = 482 m/s using the velocity coefficient
V_2 = (0.9) × (482) = 434 m/s
The absolute velocity at the inlet to the second row of moving blades, C_3, is equal to the velocity of steam leaving the fixed row of blades.
i.e., : C_3 = kC_2 = (0.9) × (316.4) = 284.8
Driving force = \dot{m} ΔC_w
For the first row of moving blades, \dot{m} ΔC_{w1} = (1) × (854) = 854 N.
For the second row of moving blades, \dot{m} ΔC_{w2} = (1) × (281.46)
N = 281.46 N
where ΔC_{w1} and ΔC_{w1} are scaled from the velocity diagram.
Total driving force = 854 + 281.46 = 1135.46 N per kg/s
Power = driving force £ blade velocity
= \frac{(1135.46) × (115)} {1000} = 130.58 kW per kg/s
Energy supplied to the wheel
= \frac{mC^2 _1} {2} =\frac{(1) × (590^2)} {(2) × (10^3)} = 174.05 kW per kg/s
Therefore, the diagram efficiency is:
η_d =\frac{(130.58) × (10^3) × (2)} {590^2} = 0.7502, or 75.02%
Maximum diagram efficiency:
= \cos^2 α_1 = \cos^2 8° = 0.9045, or 90.45%
Axial thrust on the first row of moving blades (per kg/s):
= \dot{m} (C_{a1} – C_{a2}) = (1) × (182.32 – 148.4) = 33.9 N
Axial thrust on the second row of moving blades (per kg/s):
= \dot{m} (C_{a3} – C_{a4}) = (1) × (111.3 – 97.57) = 13.73 N
Total axial thrust:
= 33.9 + 13.73 = 47.63 N per kg/s
