Question 6.DE.10: In an impulse turbine, steam is leaving the nozzle with velo......
In an impulse turbine, steam is leaving the nozzle with velocity of 950 m/s and the nozzle angle is 20°. The nozzle delivers steam at the rate of 12 kg/min. The mean blade speed is 380 m/s and the blades are symmetrical. Neglect friction losses. Calculate (1) the blade angle, (2) the tangential force on the blades, and (3) the horsepower developed.
With the help of α_1, U and C_1, the velocity triangle at the blade inlet can be constructed easily as shown in Fig. 6.27.
Applying the cosine rule to the triangle ABC,
V^2_1 = U^2 + C^2_1 – 2UC_1\cosα_1
= 950² + 380² – (2) × (950) × (380) × cos20° = 607 m/s
Now, applying the sine rule to the triangle ABC,
\frac{V_1} {\sin(α_1)} = \frac{C_1}{\sin(180° – β_1)} = \frac{C_1} {\sin(β_1)}
or:
\sin(β_1) = \frac{C_1\sin(α_1)} {V_1} = \frac{(950) × (0.342)} {607} = 0.535
so:
β_1 = 32.36°
From Triangle ACD,
C_{w1} = C_1\cos(α_1) = 950 × cos(20°) = (950) × (0.9397)
= 892.71 m/s
As β_1 = β_2, using triangle BEF and neglecting friction loss, i.e,: V_1 = V_2
BF = V_2 \cos β_2 = 607 × cos 32.36° = 512.73
Therefore,
C_{w2} = BF – U = 512.73 – 380 = 132.73 m/s
Change in velocity of whirl:
ΔC_w = C_{w1} + C_{w2} = 892.71 + 132.73 = 1025.44 m/s
Tangential force on blades:
F = \dot{m}ΔC_w = \frac{(12) × (1025.44)}{60} = 205 N
Horsepower, P = \dot{m}UΔC_w = \frac{(12) × (1025.44) × (380)} {(60) × (1000) × (0.746)} = 104.47 hp
