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The blade and fluid velocities for an axial turbine are as shown in the figure.
The magnitude of absolute velocity at entry is 300 m/s at an angle of 65° to the axial direction, while the magnitude of the absolute velocity at exit is 150 m/s. The exit velocity vector has a component in the downward direction. Given that the axial (horizontal) velocity is the same at entry and exit, the specific work (in kJ/kg) is___________ .
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- 52.88
- 51.88
- 52.81
- 52
Correct Option: C
Given: V1 = 300 m/sec
u =150 m/sec
Vf1 = Vf2
(α = 25°)
V2 = 150 m/sec
Specific work = [Vw1 + Vw2]. u
Vw1 = V1 cos 25°
Vw2 = 300 cos 25°
= 271.89 m/s
Vf1 = Vf2 = V1 sin 25
Vf2 = 300 sin 25
= 126.78 m/s
Vw2 = √Vf12 - Vf22
= √1502 - 126.782 = √6426.83
Vw2 = 80.16 m/s
Specific work = [Vw1 + Vw2]. u
= [271.89 + 80.76] × 150
= 52807.5 J/kg
= 52.81 kJ/kg