Fluid Mechanics and Hydraulic Machinery Miscellaneous
- Which combination of the following statements about steady incompressible forced vortex flow is correct?
P: Shear stress is zero at all points in the flow.
Q: Vorticity is zero at all points in the flow.
R: Velocity is directly proportional to the radius from the centre of the vortex.
S: Total mechanical energy per unit mass is constant in the entire flow field.
Select the correct answer using the codes given below:
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Clearly zero shear stress and vortex.
Correct Option: B
Clearly zero shear stress and vortex.
- The 2-D flow with, velocity
V̄ = (x + 2y + 2) î + (4 – y) ĵ is
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v̄ = (x + 2y +2)î + (4 - y)ĵ
u =x + 2y + 2, v = 4 – y∴ δv = 1, δv = - 1 δx δy ∴ δv + 1, δv = 0 δx δy
hence in compressible.Again, ω = 1 δv - δv 2 δx δy = 1 (0 - 2) = - 1. 2
hence not irrotational.Correct Option: D
v̄ = (x + 2y +2)î + (4 - y)ĵ
u =x + 2y + 2, v = 4 – y∴ δv = 1, δv = - 1 δx δy ∴ δv + 1, δv = 0 δx δy
hence in compressible.Again, ω = 1 δv - δv 2 δx δy = 1 (0 - 2) = - 1. 2
hence not irrotational.
- Existence of velocity potential implies that
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Fluid is irrotational
Correct Option: B
Fluid is irrotational
- A closed cylinder having a radius R and height H is filled with oil of density ρ. If the cylinder is rotated about its axis at an angular velocity of ω, the thrust at the bottom of the cylinder is
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We know that
δp = ρv2 = ρ.ω2r2 = δω2r δr r r
[∵ v = ω × r]p δp = r ρω2r dr 0 0 p = ρ ω2r2 2
Area of circular ring = 2πrdr
Force on elementry ring
= Intensity of pressure × Area of ring= ρ ω2r22πr dr 2
∴ Total force on the top of the cylinder= R ρ ρω2r22π r dr 0 2 = ρ ρω2r22π R r3 dr 2 0 ρ . ω22π R2 = ρ ω2 × πR4 2 4 4
Thrust at the bottom of the cylinder
= Weight of water in cylinder + Total force on the top of cylinder= ρg × πR2 × H + ρ ω2 × πR4 4 = πR2 ρω2R2 + ρgH 4 Correct Option: D
We know that
δp = ρv2 = ρ.ω2r2 = δω2r δr r r
[∵ v = ω × r]p δp = r ρω2r dr 0 0 p = ρ ω2r2 2
Area of circular ring = 2πrdr
Force on elementry ring
= Intensity of pressure × Area of ring= ρ ω2r22πr dr 2
∴ Total force on the top of the cylinder= R ρ ρω2r22π r dr 0 2 = ρ ρω2r22π R r3 dr 2 0 ρ . ω22π R2 = ρ ω2 × πR4 2 4 4
Thrust at the bottom of the cylinder
= Weight of water in cylinder + Total force on the top of cylinder= ρg × πR2 × H + ρ ω2 × πR4 4 = πR2 ρω2R2 + ρgH 4
- The horizontal and vertical hydro static forces Fx and Fy on the semicircular gate, having a width w into the plane of figure, are
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Fx = 2ρghrw
Fy = w2 - w1 πr2 . ρg × w = πr2ρgw 2 2 Correct Option: D
Fx = 2ρghrw
Fy = w2 - w1 πr2 . ρg × w = πr2ρgw 2 2