JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II MECHANICS OF FLUIDS Previous Question Paper (September - 2021 - R13)
Answer any five questions. All questions carry equal marks.
Main
Explain how you would find the resultant pressure and centre of pressure on a curved surface immersed in a liquid. Derive the same.
Answer
When a curved surface is immersed in a liquid, the pressure force exerted by the fluid varies in both magnitude and direction at every point. To find the resultant pressure force, we resolve the total force into horizontal and vertical components.
The horizontal component of the resultant force, Fx, is equal to the total pressure force on the projected area of the curved surface on a vertical plane. It acts through the centre of pressure of the projected vertical area.
Where ρ is the fluid density, g is gravity, Av is the area of the vertical projection, and h̄ is the depth of the centroid of the projected area.
The vertical component of the resultant force, Fy, is equal to the weight of the liquid supported by the curved surface up to the free surface. It acts through the centre of gravity of the volume of liquid contained above the curved surface.
Where V is the volume of the liquid column vertically above the curved surface.
The resultant force F is given by the vector sum of these components:
The angle θ that the resultant force makes with the horizontal is:
The centre of pressure (x', y') is determined by taking moments of the pressure forces about a reference axis. For a curved surface, the line of action of the resultant force passes through the point of intersection of the vertical line of action of Fy and the horizontal line of action of Fx.
Develop the expression for the relation between gauge pressure P inside a droplet of liquid and the surface tension.
Answer
Consider a small spherical droplet of liquid of radius R and surface tension σ. Let P be the gauge pressure inside the droplet. The droplet is in equilibrium under the action of pressure force and surface tension force. By splitting the droplet into two hemispheres, the forces acting on the cross-sectional area are:
- Pressure force acting on the cross-section: F_p = P × π × R2
- Surface tension force acting around the circumference: F_s = σ × 2πR
For equilibrium, these two forces must be equal:
Solving for P, we obtain the expression for gauge pressure inside a liquid droplet:
Calculate the pressure due to column of 0.3 m of water and an oil of specific gravity 0.8, and mercury of specific gravity 13.6. Take density of water as 1000 kg/m3.
Answer
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Derive the expression for continuity equation in three dimensions.
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Define the following and give practical example for each:
a) Laminar flow
b) Turbulent flow
c) Steady flow
d) Uniform flow.
Answer
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State Buckingham’s π-theorem. Why this theorem is considered superior over the Raleigh’s method for dimensional analysis?
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A horizontal venturimeter with inlet diameter 20 cm and throat diameter 10 cm is used to measure the oil of specific gravity 0.8. The discharge of oil through venturimeter is 60 litres/s. Find the reading of oil-mercury differential manometer. Take Cd = 0.98.
Answer
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What is Euler’s equation of motion? How will you obtain Bernoulli’s equation from it?
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A pitot tube is inserted in a pipe of 300 mm diameter. The static pressure in pipe is 100 mm of mercury (vacuum). The stagnation pressure at the centre of the pipe, recorded by the pitot-tube is 0.981 N/cm2. Calculate the rate of flow of water through pipe, if the mean velocity of flow is 0.85 times the central velocity. Take Cv = 0.98.
Answer
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Find displacement thickness and the momentum thickness for the velocity distribution in the boundary layer given by:
Where:
u = Velocity in boundary layer at a distance y
U = Free stream velocity
δ = Boundary layer thickness.
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Derive an expression for the loss of head due to sudden enlargement.
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Derive an expression for the loss of head due to sudden contraction of a pipe.
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