Fluid Mechanics and Hydraulic Machinery Miscellaneous
 Water flows through a pipe of diameter 0.30 m. What would be the velocity V for the conditions shown in the figure below?

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h = x 1 − ρ_{mano} ρ_{pipe} = 0.3 1 − 800 1000
= 0.06 m of water
V_{1} = √2gh = √2 × 9.8 × 0.6
= 1.084 m/sCorrect Option: D
h = x 1 − ρ_{mano} ρ_{pipe} = 0.3 1 − 800 1000
= 0.06 m of water
V_{1} = √2gh = √2 × 9.8 × 0.6
= 1.084 m/s
 In a hand operated liquid sprayer (figure shown below) the liquid from the container rises to the top of the tube because of

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NA
Correct Option: C
NA
 An incompressible fluid (kinematic viscosity, 7.4 × 10^{–7} m²/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates: the shear stress in Pascals on the surface of top plate is

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Given:
υ = 7.4 × 10^{–7} m^{2} /s
ρ = 0.88 × 1000 = 880
v = 0.5 m/sShear stress, τ = μ⋅ du = (υ × ρ) × du dy dy = 7.4 × 10^{–7} × 800 × 0.5 = 0.6512. (0.0005) Correct Option: B
Given:
υ = 7.4 × 10^{–7} m^{2} /s
ρ = 0.88 × 1000 = 880
v = 0.5 m/sShear stress, τ = μ⋅ du = (υ × ρ) × du dy dy = 7.4 × 10^{–7} × 800 × 0.5 = 0.6512. (0.0005)
 The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u = u_{0} (1– 4r²/D²), where r is the radial distance from the center. If the viscosity of the fluid is ρ, the pressure drop across a length L of the pipe is

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By Hagen – Poiseuille law, for steady laminar flow in circular pipes
τ = –μ ∂u ∂r τ = –∂p . r ∂x 2 μ ∂u = ∂P ⋅ r ∂r ∂x 2 μu_{0} −8r = P . r D² L 2 .... ∵ u = u_{0} 1 − 4r² D² = –16μLu_{0} D²
[(–) sign is due to drop]Correct Option: D
By Hagen – Poiseuille law, for steady laminar flow in circular pipes
τ = –μ ∂u ∂r τ = –∂p . r ∂x 2 μ ∂u = ∂P ⋅ r ∂r ∂x 2 μu_{0} −8r = P . r D² L 2 .... ∵ u = u_{0} 1 − 4r² D² = –16μLu_{0} D²
[(–) sign is due to drop]
 Consider steady laminar incompressible axisymmetric fully developed viscous flow through a straight circular pipe of constant crosssectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is

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Reynold’s number = Inertia force = 5 Viscous force Correct Option: A
Reynold’s number = Inertia force = 5 Viscous force