Fluid Mechanics and Hydraulic Machinery Miscellaneous
- 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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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/kgCorrect 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
- Consider two hydraulic turbines having identical specific speed and effective head at the inlet. If the speed ratio (N1 /N2) of the two turbines is 2, then the respective power ratio (P1 /P2) is ____.
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Ns N√P H5/4
Given Ns1 = Ns2, H1 H2N1 = 2, P1 = ? N2 P2
N1 √P1 = N2√P2⇒ P1 = 
N1 
2 P2 N2 = 
1 
2 = 0.25 2 Correct Option: C
Ns N√P H5/4
Given Ns1 = Ns2, H1 H2N1 = 2, P1 = ? N2 P2
N1 √P1 = N2√P2⇒ P1 = N1 2 P2 N2 = 1 2 = 0.25 2
- Which of the following statements are TRUE, when the cavitation parameter σ = 0?
i. the local pressure is reduced to vapor pressure
ii. cavitation starts
iii. boiling of liquid starts
iv. cavitation stops
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σ = 0 implies (i), (ii) and (iii)
Correct Option: D
σ = 0 implies (i), (ii) and (iii)
- At the inlet of an axial impulse turbine rotor, the blade linear speed is 25 m/s, the magnitude of absolute velocity is 100 m/s and the angle between them is 25°. The relative velocity and the axial component of velocity remain the same between the inlet and outlet of the blades. The blade inlet and outlet velocity triangles are shown in figure. Assuming no losses, the specific work (in J/kg) is _______.
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Given α = 25º
Vr1 sin β1 =V1 sinα
∴ 78 sin β1 = 100 sin 25º
∴ β1 = 32.81º
∵ Vr1 = Vr2
∴ β1 = β2 = 32.81º
Now ∴ ∆Vw = Vr1 cosβ1 + Vr2 cosβ2
= 2Vr1 cosβ1
= 2 × 78 × cos32.81°
∴ ∆Vw = 131.1136 m/sec
∴ Sp. power = ∆Vw . u = 131.1136 × 25
= 3277.84 J/kg.Correct Option: A
Given α = 25º
Vr1 sin β1 =V1 sinα
∴ 78 sin β1 = 100 sin 25º
∴ β1 = 32.81º
∵ Vr1 = Vr2
∴ β1 = β2 = 32.81º
Now ∴ ∆Vw = Vr1 cosβ1 + Vr2 cosβ2
= 2Vr1 cosβ1
= 2 × 78 × cos32.81°
∴ ∆Vw = 131.1136 m/sec
∴ Sp. power = ∆Vw . u = 131.1136 × 25
= 3277.84 J/kg.
- A siphon draws water from a reservoir and discharge it out at atmospheric pressure.
Assuming ideal fluid and the reservoir is large, the velocity at point Pin the siphon tube is
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By energy conservation, velocity at point
Q = √2g(h2 - h1)
As there is a continuous and uniform flow, so velocity of liquid at point Q and P is same.
Vp =√2g(h2 - h1)Correct Option: B
By energy conservation, velocity at point
Q = √2g(h2 - h1)
As there is a continuous and uniform flow, so velocity of liquid at point Q and P is same.
Vp =√2g(h2 - h1)
