JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II FLUID MECHANICS AND OPEN CHANNEL HYDRAULICS Previous Question Paper (March - 2021 - R13)

B.Tech | II | I | R13 | March - 2021

Answer PaperPaper Code: 113AGMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

Derive an expression for center of pressure of an inclined immersed surface.

15 marks

Answer

Consider a plane surface of arbitrary shape immersed in a liquid of specific weight w, inclined at an angle θ with the free surface of the liquid. Let A be the total area of the surface, G be the center of gravity, and P be the center of pressure. Let h_G be the depth of the center of gravity and h_P be the depth of the center of pressure from the free surface. Let y_G and y_P be the distances of the center of gravity and center of pressure from the axis O-O (the intersection of the plane of the surface and the free surface).

The total pressure force F acting on the surface is given by:

F = w A h_G = w A y_G sin θ

The moment of this force about the axis O-O must be equal to the sum of the moments of the elementary forces acting on the surface. Taking an elemental area dA at a distance y from O-O, the pressure force on this element is dF = w(y sin θ)dA. The moment of this force about O-O is dM = dF * y = w y^2 sin θ dA.

Integrating over the entire area, the total moment M is:

M = ∫ w y^2 sin θ dA = w sin θ ∫ y^2 dA

Since ∫ y^2 dA is the moment of inertia of the surface about the axis O-O (denoted as I_O), we have M = w sin θ I_O. By the principle of moments, the total force F acting at the center of pressure y_P must produce the same moment:

F * y_P = w sin θ I_O

Substituting F = w A y_G sin θ:

(w A y_G sin θ) * y_P = w sin θ I_O

Simplifying for y_P:

y_P = I_O / (A y_G)

Using the parallel axis theorem, I_O = I_G + A y_G^2, where I_G is the moment of inertia about the centroidal axis:

y_P = (I_G + A y_G^2) / (A y_G) = y_G + I_G / (A y_G)

Since h_P = y_P sin θ and h_G = y_G sin θ, the depth of the center of pressure is:

h_P = h_G + (I_G sin^2 θ) / (A h_G)
2

Explain about the following:

15 marks
a

Bourdon’s pressure gauge

8 marks

Answer

A Bourdon pressure gauge is a mechanical device used to measure gauge pressure. It consists of a thin-walled, hollow, elastic metal tube with an elliptical cross-section, which is bent into a circular arc known as a Bourdon tube. One end of the tube is fixed and connected to the fluid source, while the other end is free to move.

When fluid under pressure enters the tube, the cross-section tends to become circular, causing the tube to straighten out. This small movement of the free end is amplified through a mechanical linkage system consisting of a sector gear and a pinion. The pinion rotates a pointer over a calibrated circular scale, providing a direct reading of the pressure. It is widely used due to its simplicity, robustness, and ability to measure a wide range of pressures.

b

Different flow lines

7 marks

Answer

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3

Explain Prandtl’s experiment of boundary layer separation.

15 marks

Answer

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4

Derive Bernoulli’s equation from the fundamentals.

15 marks

Answer

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5

What is velocity of approach? Explain how it is determined.

15 marks

Answer

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6

Discuss in detail, the classifications of notches and weirs along with their discharge formulae.

15 marks

Answer

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7

Describe different types of open channels and different types of flows in open channels.

15 marks

Answer

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8

Derive the condition for maximum velocity in respect of circular channel section.

15 marks

Answer

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About this paper

This page contains the question paper of FLUID MECHANICS AND OPEN CHANNEL HYDRAULICS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech AG II I, March - 2021, R13.

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