## Fluid Mechanics and Hydraulic Machinery Miscellaneous

#### Fluid Mechanics and Hydraulic Machinery

1. 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

1. 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

##### Correct 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

1. 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/m3.

1. Applying Bernoulli’s equation between sections (i) & A

 P1 = PA + VA² ρg 2g

 PA - P1 = -VA² ρg 2g

P1 = Pambient
∴ PA - P1 = (PA)gauge
 ∴ (PA)gauge == -VA² ρg 2g

 (PA)gauge = -1 × 1² = 0.5N/m² 2

##### Correct Option: A

Applying Bernoulli’s equation between sections (i) & A

 P1 = PA + VA² ρg 2g

 PA - P1 = -VA² ρg 2g

P1 = Pambient
∴ PA - P1 = (PA)gauge
 ∴ (PA)gauge == -VA² ρg 2g

 (PA)gauge = -1 × 1² = 0.5N/m² 2

1. A two-dimensional 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

1. NA

##### Correct Option: D

NA

1. 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

1. NA

##### Correct Option: C

NA

1. For a two-dimensional 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
3. y-component of acceleration,
 ay = - y (x² + y²)²

4. x-component of acceleration,
 ax = -(x + y) (x² + y²)²

1.  axu δu + v δu δx δy

 = x(x² + y² -2x²)-2xy² = - x3 - xy² (x² + y²)(x² + y²) (x² + y²)

 ∴ ax = x (x² + y²)²

 ay = u δv + v δv δx δy

 = -x² + yx² - y3 = - 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

 axu δu + v δu δx δy

 = x(x² + y² -2x²)-2xy² = - x3 - xy² (x² + y²)(x² + y²) (x² + y²)

 ∴ ax = x (x² + y²)²

 ay = u δv + v δv δx δy

 = -x² + yx² - y3 = - 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.