Question 6.8: In one stage of an impulse turbine the velocity of steam at ......
In one stage of an impulse turbine the velocity of steam at the exit from the nozzle is 460 m/s, the nozzle angle is 22° and the blade angle is 33°. Find the blade speed so that the steam shall pass on without shock. Also find the stage efficiency and end thrust on the shaft, assuming velocity coefficient = 0.75, and blades are symmetrical.
From triangle ABC (Fig. 6.25):
C_{w1} = C_1 cos 22° = 460 cos 22° = 426.5 m/s
and:
C_{a1} = C_1 sin 22° = 460 sin 22° = 172.32 m/s
Now, from triangle BCD:
BD = \frac{C_{a1}} {\tan (33°)} = \frac{172.32} {0.649} = 265.5
Hence, blade speed is given by:
U = C_{w1} – BD = 426.5 – 265.5 = 161 m/s
From Triangle BCD, relative velocity at blade inlet is given by:
V_1 = \frac{C_{a1}} {\sin (33°)} = \frac{172.32} {0.545} = 316.2 m/s
Velocity coefficient:
k = \frac{V_2} {V_1}, \text{or} V_2 = kV_1 = (0.75) × (316.2) = 237.2 m/s
From Triangle BEF,
BF = V_2 cos (33°) = 237.2 × cos (33°) = 198.9
and
C_{w2} = AF = BF – U = 198.9 – 161 = 37.9 m/s
C_{a2} = V_2 sin (33°) = 237.2 sin (33°) = 129.2 m/s
The change in velocity of whirl:
ΔC_w = C_{w1} + C_{w2} = 426.5 + 37.9 = 464.4 m/s
Diagram efficiency:
η_d = \frac{2UΔC_w} {C^2_1} = \frac{(2) × (464.4) × (161)} {460^2} = 0.7067, or 70.67%.
End thrust on the shaft per unit mass flow:
C_{a1} – C_{a2} = 172.32 – 129.2 = 43.12 N
