Question 6.DE.11: In an impulse turbine, the velocity of steam at the exit fro......
In an impulse turbine, the velocity of steam at the exit from the nozzle is 700 m/s and the nozzles are inclined at 22° to the blades, whose tips are both 34°. If the relative velocity of steam to the blade is reduced by 10% while passing through the blade ring, calculate the blade speed, end thrust on the shaft, and efficiency when the turbine develops 1600 kW.
Velocity triangles for this problem are shown in Fig. 6.28.
From the triangle ACD,
C_{a1} = C_1 \sinα_1 = 700 × sin 22° = 262.224 m/s
and
V_1 = \frac{C_{a1}} {\sin (β_1)} = \frac{262.224} {\sin 34°} = 469.32 m/s
Whirl component of C_1 is given by
C_{w1} = C_1 cos(α_1) = 700 cos(22°) = 700 × 0.927 = 649 m/s
Now, BD = C_{w1} – U = V_1 \cosβ_1 = (469.32) × (0.829) = 389
Hence, blade speed
U = 649 – 389 = 260 m/s
Using the velocity coefficient to find V_2:
i:e:; V_2 = (0.90) × (469.32) = 422.39 m/s
From velocity triangle BEF,
C_{a2} = V_2\sin (β_2) = 422.39 sin 34° = 236.2 m/s
And
U + C_{w2} = V_2 cos 34° = (422.39) × (0.829) = 350.2 m/s
Therefore,
C_{w2} = 350.2 – 260 = 90.2 m/s
Then,
ΔC_w = C_{w1} + C_{w2} = 649 + 90.2 = 739.2 m/s
Mass flow rate is given by:
P = \dot{m}UΔC_w
or
\dot{m} = \frac{(1600) × (1000)} {(739.2) × (260)} = 8.325 kg/s
Thrust on the shaft,
F = \dot{m}(C_{a1} – C_{a2}) = 8.325(262.224 – 236.2) = 216.65 N
Diagram efficiency:
η_d = \frac{2UΔC_w} {C^2_1} = \frac{(2) × (739.2) × (260)} {700^2} = 0.7844, or 78.44%.
