Fluid Mechanics and Hydraulic Machinery Miscellaneous
 A water container is kept on a weighing balance. Water from a tap is falling vertically into the container with a volume flow rate of Q; the velocity of the water when it hits the water surface is U. At a particular instant of time the total mass of the container and water is m. The force registered by the weighing balance at this instant of time is

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Volume flow rate = Q
Mass of water strike = ρQ
Velocity of the water when it hit the water surface = U
Force on weighing balance due to water strike = Initial momentum – final momentum
= ρQU – 0= ρQU
(since final velocity is perpendicular to initial velocity)
Now total force on weighing balance = mg + ρQUCorrect Option: A
Volume flow rate = Q
Mass of water strike = ρQ
Velocity of the water when it hit the water surface = U
Force on weighing balance due to water strike = Initial momentum – final momentum
= ρQU – 0= ρQU
(since final velocity is perpendicular to initial velocity)
Now total force on weighing balance = mg + ρQU
 A fan in the duct shown below sucks air from the ambient and expels it as a jet at 1 m/s to the ambient. Determine the gauge pressure at the point marked as A. Take the density of air as 1 kg/m^{3}.

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Applying Bernoulli’s equation between sections (i) & A
P_{1} = P_{A} + V_{A}² ρg 2g P_{A}  P_{1} = V_{A}² ρg 2g
P_{1} = Pambient
∴ P_{A}  P_{1} = (P_{A})gauge∴ (P_{A})gauge == V_{A}² ρg 2g (P_{A})gauge = 1 × 1² = 0.5N/m² 2
Correct Option: A
Applying Bernoulli’s equation between sections (i) & A
P_{1} = P_{A} + V_{A}² ρg 2g P_{A}  P_{1} = V_{A}² ρg 2g
P_{1} = Pambient
∴ P_{A}  P_{1} = (P_{A})gauge∴ (P_{A})gauge == V_{A}² ρg 2g (P_{A})gauge = 1 × 1² = 0.5N/m² 2
 A twodimensional incompressible frictionless flow field is given by μ = xi  yj. If ρ is the density of the fluid, the expression for pressure gradient vector at any point in the flow field is given as

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NA
Correct Option: D
NA
 For a steady flow, the velocity field is V = (–x² + 3y) i + (2xy)j. The magnitude of the acceleration of a particle at (1,  1) is

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NA
Correct Option: C
NA
 For a twodimensional flow, the velocity field is
u = x i + y j x² + y² x² + y²
where i and j are the basis vectors in the x – y Cartesian coordinate system. Identify the CORRECT statements from below.
1. The flow is incompressible
2. The flow is unsteady
3. ycomponent of acceleration,a_{y} =  y (x² + y²)²
4. xcomponent of acceleration,a_{x} = (x + y) (x² + y²)²

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a_{x}u δu + v δu δx δy = x(x² + y² 2x²)2xy² =  x^{3}  xy² (x² + y²)(x² + y²) (x² + y²) ∴ a_{x} = x (x² + y²)² a_{y} = u δv + v δv δx δy = x² + yx²  y^{3} =  y (x² + y²)^{3} (x² + y²)²
The velocity components are not functions of time, so flow is steady according to continuity equation,
Since it satisfies the above continuity equation for 2D and incompressible flow.
∴ The flow is incompressible.Correct Option: B
a_{x}u δu + v δu δx δy = x(x² + y² 2x²)2xy² =  x^{3}  xy² (x² + y²)(x² + y²) (x² + y²) ∴ a_{x} = x (x² + y²)² a_{y} = u δv + v δv δx δy = x² + yx²  y^{3} =  y (x² + y²)^{3} (x² + y²)²
The velocity components are not functions of time, so flow is steady according to continuity equation,
Since it satisfies the above continuity equation for 2D and incompressible flow.
∴ The flow is incompressible.