Fluid Mechanics and Hydraulic Machinery Miscellaneous
- Velocity vector of a flow field is given as = 2xyî - x²zĵ. The vorticity vector at (1, 1, 1) is
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v = 2xyi – x 2zj
Velocity of a vector = ∇ × v= î 0 - δ ( - x²z) - î 0 - δ (2xy) - k̂ 0 - δ (-x²z) - δ (2xy) δz δz δx δy
= î(-x²) + k̂(-2 × z - 2X)
At(1 , 1 , 1),
∇ × v = î + k̂ (-2 -2) = i - 4kCorrect Option: D
v = 2xyi – x 2zj
Velocity of a vector = ∇ × v= î 0 - δ ( - x²z) - î 0 - δ (2xy) - k̂ 0 - δ (-x²z) - δ (2xy) δz δz δx δy
= î(-x²) + k̂(-2 × z - 2X)
At(1 , 1 , 1),
∇ × v = î + k̂ (-2 -2) = i - 4k
- A Prandtl tube (Pitot-static tube with C = 1) is used to measure the velocity of water. The differential manometer reading is 10 mm of liquid coiumn with a relative density of 10. Assuming g = 9.8 m/s², the velocity of water (in m/s) is _____ .
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Velocity as water = Cv √2gh
Cv = 1 (Given)h = x sg - 1 s0
= 0.01 (10 – 1) = 0.09 m
∴ velocity of flow = √2 × 9.8 × 0.09 = 1.328 m/ sCorrect Option: C
Velocity as water = Cv √2gh
Cv = 1 (Given)h = x sg - 1 s0
= 0.01 (10 – 1) = 0.09 m
∴ velocity of flow = √2 × 9.8 × 0.09 = 1.328 m/ s
- A streamline and an equipotential line in a flow field
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dy × dy = - 1 dx φ dx ψ
Slope of equipotential Line × slope of stream function = – 1
They are orthogonal to each other.Correct Option: B
dy × dy = - 1 dx φ dx ψ
Slope of equipotential Line × slope of stream function = – 1
They are orthogonal to each other.
- The arrangement shown in the figure measures the velocity V of a gas of density 1 kg/m³ flowing through a pipe. The acceleration due to gravity is 9.81 m/s². If the manometric fluid is water (density 1000 kg/m³) and the velocity V is 20 m/ s, the differential head h(in mm) between the two arms of the manometer is _________.
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Dynamic pressure of gas = (δgh)water
1 δgasv2 = δm × 9.81 × h 2 1 × 1 × (20)² = 1000 × 9.81 × h 2
h = 0.02038 m of water
h = 20.38 mm of waterCorrect Option: D
Dynamic pressure of gas = (δgh)water
1 δgasv2 = δm × 9.81 × h 2 1 × 1 × (20)² = 1000 × 9.81 × h 2
h = 0.02038 m of water
h = 20.38 mm of water
- The Newtonian fluid has the following velocity field:
= x²yî + 2xy²ĵ - yz³k̂
The rate shear deformation εyz at the point x = –2, y = –1 and z = 2 for the given flow is
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V = x⊃î + 2xy²zĵ - yz²̂k
Shear strainyz = 1 δv + δw = 1 (2xy² - z³) 2 δz δy 2
at (–2π, 2), we getyz = 1 (- 4 - 8)) = - 6 2 Correct Option: A
V = x⊃î + 2xy²zĵ - yz²̂k
Shear strainyz = 1 δv + δw = 1 (2xy² - z³) 2 δz δy 2
at (–2π, 2), we getyz = 1 (- 4 - 8)) = - 6 2