Fluid Mechanics and Hydraulic Machinery Miscellaneous
Direction: The gap between a moving circular plate and a stationary surface is being continuously reduced, as the circular plate comes down at a uniform speed V towards the stationary bottom surface, as shown in the figure. In the process, the fluid contained between the two plates flows out radially. The fluid is assumed to be incompressible and inviscid.
- The radial velocity Vr at any radius r, when the gap width is h, is
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V × πr² = Vr × 2πrh
Vr = Vr 2h
Correct Option: A
V × πr² = Vr × 2πrh
Vr = Vr 2h
- In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by v = u0 (1 + 3x/L)i, where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is
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v = u0 1 + 3x L ∴ dx = u0 1 + 3x dt L ⇒ dx = dt u0 1 + 3x L ⇒ T = L ln(4) = L ln(4) 3u0 3u0
Correct Option: B
v = u0 1 + 3x L ∴ dx = u0 1 + 3x dt L ⇒ dx = dt u0 1 + 3x L ⇒ T = L ln(4) = L ln(4) 3u0 3u0
- A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of 120 m from the centre of the whirlpool. The whirlpool can be described by the following velocity distribution:
Vr = - 60 × 103 m/s and Vθ = 300 × 103 m/s 2πr 2πr
where r (in meters) is the distance from the centre of the whirlpool. What will be the distance of the leaf from the centre when it has moved through half a revolution?
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Radial distance = 120 m
Vr = - 60 × 103 m /s 2πr & Vθ = 300 × 103 m /s 2πr
& θ = πVr = -1 Vθ 5
we know thatVr = dr ...(i) dt & Vθ = rw = r dθ dt -5Vr = r dθ dt Vr = -r × dθ ...(ii) 5 dt
By equating (i) & (ii), we getdr = -r . dθ dt 5 dt
Integrating both sides, we getln r = -π 120 5
By solving above, we get
r = 64 mCorrect Option: B
Radial distance = 120 m
Vr = - 60 × 103 m /s 2πr & Vθ = 300 × 103 m /s 2πr
& θ = πVr = -1 Vθ 5
we know thatVr = dr ...(i) dt & Vθ = rw = r dθ dt -5Vr = r dθ dt Vr = -r × dθ ...(ii) 5 dt
By equating (i) & (ii), we getdr = -r . dθ dt 5 dt
Integrating both sides, we getln r = -π 120 5
By solving above, we get
r = 64 m
- The velocity components in the x and y directions of a two dimensional potential flow are u and v, respectively.
Then , ∂u is equal to ∂x
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∂²φ + ∂²φ = 0 ...(i) (Laplace Equation should be satisfied) ∂x² ∂y² ∴ u = - ∂φ and v = - ∂φ ∂x ∂y ⇒ ∂u = - ∂²φ and ∂v = - ∂²φ ∂x ∂x² ∂y ∂y²
Putting values in equation (i), we get-∂u + -∂v = 0 ∂x ∂y ⇒ ∂u = - ∂v ∂x ∂y
Correct Option: D
∂²φ + ∂²φ = 0 ...(i) (Laplace Equation should be satisfied) ∂x² ∂y² ∴ u = - ∂φ and v = - ∂φ ∂x ∂y ⇒ ∂u = - ∂²φ and ∂v = - ∂²φ ∂x ∂x² ∂y ∂y²
Putting values in equation (i), we get-∂u + -∂v = 0 ∂x ∂y ⇒ ∂u = - ∂v ∂x ∂y
- The velocity components in the x and y directions are given by u = λxy3 – x2y, v = xy2 – ( 3 / 4 )y4. The value of λ for a possible flow field involving an incompressible fluid is
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u = λxy3 – x2y ; ∂u = λy3 – 2xy ∂x v = xy2 – 3 y4 ; ∂v = 2xy - 3y3 4 ∂y
For 2D incompressible flow,∂u + ∂v = 0 ∂x ∂y
⇒ xy3 – 2xy + 2xy - 3y3 = 0
λ = 3
Correct Option: D
u = λxy3 – x2y ; ∂u = λy3 – 2xy ∂x v = xy2 – 3 y4 ; ∂v = 2xy - 3y3 4 ∂y
For 2D incompressible flow,∂u + ∂v = 0 ∂x ∂y
⇒ xy3 – 2xy + 2xy - 3y3 = 0
λ = 3