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JNTU B.Tech II MECHANICS OF FLUIDS Previous Question Paper (March - 2021 - R15)

B.Tech | II | I | R15 | March - 2021

Answer PaperPaper Code: 123AD/113ADMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Define the terms: i) Density ii) Specific gravity iii) Surface tension.

7 marks

Answer

i) Density (Mass Density): Defined as the ratio of the mass of a fluid to its volume. It is denoted by ρ and has units of kg/m3.

ii) Specific Gravity: Defined as the ratio of the density of a fluid to the density of a standard fluid (water at 4°C for liquids). It is a dimensionless quantity.

iii) Surface Tension: Defined as the tensile force per unit length acting on the surface of a liquid in contact with a gas or another liquid, caused by molecular cohesion. It is denoted by σ and has units of N/m.

b

Calculate the capillary effect in millimeters in a glass tube of 4mm diameter, when immersed in i) water, and ii) mercury. The temperature of the liquid is 20°C and the value of the surface tension of water and mercury at 20°C in contact with air are 0.073575 N/m and 0.51 N/m respectively. The angle of contact for water is zero and that for mercury is 130°.

8 marks

Answer

The capillary rise or fall is given by the formula:

h = (4σ cos θ) / (ρ g d)

Given: Diameter d = 0.004 m, g = 9.81 m/s2. For water: σ = 0.073575 N/m, θ = 0°, ρ = 1000 kg/m3. For mercury: σ = 0.51 N/m, θ = 130°, ρ = 13.6 × 1000 kg/m3.

  1. Water: h = (4 × 0.073575 × cos 0°) / (1000 × 9.81 × 0.004) = 0.2943 / 39.24 = 0.0075 m = 7.5 mm (rise).
  2. Mercury: h = (4 × 0.51 × cos 130°) / (13600 × 9.81 × 0.004) = (2.04 × -0.6428) / 533.664 = -0.002457 m = -2.457 mm.

The negative sign indicates a capillary depression of 2.457 mm for mercury.

215 marks
a

Define the equation of continuity. Obtain an expression for continuity equation for a three dimensional flow.

7 marks

Answer

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b

Define the terms: i) Velocity potential function ii) Stream function.

8 marks

Answer

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3

A horizontal venturimeter with inlet diameter 20cm and throat diameter 10cm is used to measure the flow of oil of specific gravity 0.8. The discharge of oil through venturimeter is 60 liters/s. Find the reading of the oil-mercury differential manometer. Take Cd=0.98.

15 marks

Answer

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415 marks
a

State Buckingham’s π-theorem.

7 marks

Answer

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b

The efficiency η of a fan depends on density ρ, dynamic viscosity μ of the fluid, angular velocity ω, diameter D of the rotor and the discharge Q. Express η in terms of dimensional parameters.

8 marks

Answer

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5

Explain the following: a) Laminar boundary layer b) Turbulent boundary layer c) Laminar sub-layer d) Boundary layer thickness.

15 marks

Answer

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6

Calculate the discharge through a pipe of diameter 200mm when the difference of pressure head between the two ends of a pipe 500m apart is 4m of water. Take the value of f=0.009 in the formula

hf = 4fLv2 / d2g

15 marks

Answer

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7

A main pipe divides into two parallel pipes which again forms one pipe. The length and diameter for the first parallel pipe are 200m and 1.0m respectively, while the length and diameter of the 2nd parallel pipe are 200m and 0.8m. Find the rate of flow in each parallel pipe, if total flow in the main is 3.0 m3/s. The coefficient of friction for each parallel pipe is same and equal to 0.005.

15 marks

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8

A rectangular plane surface is 2m wide and 3m deep. It lies in vertical plane in water. Determine the total pressure and position of center of pressure on the plane surface when its upper edge is horizontal and a) Coincides with water surface b) 2.5m below the free water surface.

15 marks

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