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EAT 213/4 FLUID MECHANICS & HYDRAULICS

EAT 213/4 FLUID MECHANICS & HYDRAULICS. CHAPTER 1 INTRODUCTION ( PROPERTIES OF FLUID). Course Outcome. CO1 Ability to understand and apply the basic concept of fluid mechanics. Contents. 1.0 Introduction 1.1 Density and Specific Gravity, Specific Weight 1.2 Viscosity

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EAT 213/4 FLUID MECHANICS & HYDRAULICS

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  1. EAT 213/4 FLUID MECHANICS & HYDRAULICS CHAPTER 1 INTRODUCTION (PROPERTIES OF FLUID)

  2. Course Outcome CO1 Ability to understand and apply the basic concept of fluid mechanics.

  3. Contents 1.0 Introduction 1.1 Density and Specific Gravity, Specific Weight 1.2 Viscosity 1.3 Surface Tension 1.4 Newtonian & Non-Newtonian Fluids

  4. Introduction • This chapter will begin with several concepts, definition, terminologies and approaches which should be understood by the students. • Then, it introduces the student with typical properties of fluid which are then being used extensively in the next chapters for example pressure, velocity, density and viscosity. • Some of these can be used to classify type and characteristic of fluid, such as whether a fluid is incompressible or not or whether the fluid is Newtonian or non-Newtonian.

  5. Introduction Fluid mechanics is a division in applied mechanics related to the behaviour of liquid or gas which is either in rest or in motion. The study related to a fluid in rest or stationary is referred to fluid static, otherwise it is referred to as fluid dynamic. Fluid can be defined as a substance which can deform continuously when being subjected to shear stress at any magnitude. In other words, it can flow continuously as a result of shearing action. This includes any liquid or gas.

  6. Cont.. Thus, with exception to solids, any other matters can be categorised as fluid. In microscopic point of view, this concept corresponds to loose or very loose bonding between molecules of liquid or gas, respectively. Examples of typical fluid used in engineering applications are water, oil and air. An analogy of how to understand different bonding in solids and fluids is depicted in Fig. 1.1

  7. Figure 1.1

  8. For solid, imagine that the molecules can be fictitiously linked to each other with springs. In fluid, the molecules can move freely but are constrained through a traction force called cohesion.This force is interchangeable from one molecule to another.For gases, it is very weak which enables the gas to disintegrate and move away from its container. For liquids, it is stronger which is sufficient enough to hold the molecule together and can withstand high compression, which is suitable for application as hydraulic fluid such as oil. On the surface, the cohesion forms a resultant force directed into the liquid region and the combination of cohesion forces between adjacent molecules from a tensioned membrane known as free surface.

  9. Density, ρ Definition: mass per unit volume, Densityρ slightly affected by changes in temperature and pressure.  = mass/volume = m/V Units: kg/m3 Typical values: Water = 1000 kg/m3; Air = 1.23 kg/m3

  10. Density of Ideal Gas Equation of state: An equation that relates the pressure, temperature and density (or specific volume) of a substance. Ideal gas equation: PV = nRT Where, R = 8.314 J/mol.K = 0.287 kPa.m3/kg.K Question: Write down the equation above in term of density.

  11. Specific Volume, ϒ Specific Volume is volume per unit mass. Given, V = 12 m3 m = 3 kg Then, ρ = m/V = 3/12 = 0.25 kg/m3 ϒ = 1/ρ = 4 m3/kg

  12. Specific Weight  Definition: weight of the fluid per unit volume Arising from the existence of a gravitational force, The relationshipand g can be found using the following: Since  = m/V therefore  = g Units: N/m3 Typical values: Water = 9814 N/m3; Air = 12.07 N/m3

  13. Specific Gravity (S.G.) The specific gravity (or relative density) can be defined in two ways: Definition 1: A ratio of the density of a liquid to the density of water at standard temperature and pressure (STP) (20 C, 1 atm), or Definition 2: A ratio of the specific weight of a liquid to the specific weight of water at standard temperature and pressure (STP) (20C, 1 atm), Unit: ?

  14. Example 1.1 A reservoir of oil has a mass of 825 kg. The reservoir has a volume of 0.917 m3. Compute the density, specific weight, and specific gravity of the oil. Solution:

  15. Example 1.2 Determine the density, specific gravity and mass of the air in a room whose dimensions are 4 x 5 x 6 m at 100 kPa and 25 °C?

  16. Example 1.2 (Solution)

  17. Viscosity Viscosity, , is a measure of resistance to fluid flow as a result of intermolecular cohesion. In other words, viscosity can be seen as internal friction to fluid motion which can then lead to energy loss. Different fluids deform at different rates under the same shear stress. The ease with which a fluid pours is an indication of its viscosity. Fluid with a high viscosity such as syrup deforms more slowly than fluid with a low viscosity such as water. The viscosity is also known as dynamic viscosity. Units: N.s/m2 or kg/ms Typical values: Water = 1.14x10-3 kg/ms; Air = 1.78x10-5 kg/ms

  18. Kinematic Viscosity Definition: is the ratio of the viscosity to the density; will be found to be important in cases in which significant viscous and gravitational forces exist. Units: m2/s Typical values: Water = 1.14x10-6 m2/s; Air = 1.46x10-5 m2/s; In general, viscosity of liquids with temperature, whereas viscosity of gases with in temperature.

  19. Surface Tension Surface tension coefficient s can be defined as the intensity of intermolecular traction per unit length along the free surface of a fluid, and its SI unit is N/m. The surface tension effect is caused by unbalanced cohesion forces at fluid surfaces which produce a downward resultant force which can physically seen as a membrane. The coefficient is inversely proportional to temperature and is also dependent on the type of the solid interface. For example, a drop of water on a glass surface will have a different coefficient from the similar amount of water on a wood surface.

  20. Surface Tension • Liquid droplets behave like small spherical balloons filled with liquid, and the surface of the liquid acts like stretched elastic membrane under tension. • The pulling force that causes this is: • - due to the attractive forces between molecules • Called surface tension. • Attractive force on surface molecule is not symmetric. • A repulsive forces from interior molecules causes the liquid to minimize its surface area and attain a spherical shape. XNNXN

  21. Newtonian and Non-Newtonian Fluid obey refer Newton’s law of viscosity Fluid Newtonian fluids Example: Air Water Oil Gasoline Alcohol Kerosene Benzene Glycerine Newton’s’ law of viscosity is given by; (1.1)  = shear stress  = viscosity of fluid du/dy = shear rate, rate of strain or velocity gradient • The viscosity  is a function only of the condition of the fluid, particularly its temperature. • The magnitude of the velocity gradient (du/dy) has no effect on the magnitude of .

  22. Do not obey Non- Newtonian fluids Newton’s law of viscosity Fluid Newtonian and Non-Newtonian Fluid • The viscosity of the non-Newtonian fluid is dependent on the velocity gradientas well as the condition of the fluid. • Newtonian Fluids • a linear relationship between shear stress and the velocity gradient (rate of shear), • the slope is constant • the viscosity is constant • non-Newtonian fluids • slope of the curves for non-Newtonian fluids varies

  23. If the gradient m is constant, the fluid is termed as Newtonian fluid. Otherwise, it is known as non-Newtonian fluid. Fig. 1.5 shows several Newtonian and non-Newtonian fluids.

  24. Summary of Chapter 1 • Introduction of Fluid Mechanics (Reading) • Calculation: Properties of fluid;ρ, , γ, S.G Viscosity, μ andѵ • Surface Tension, Newtonian & Non-Newtonian fluid (Reading)

  25. Example 1.3 (Tips):Test 1 (Academic Session 2014/15, Sem 1) Q1 (a) : Calculation from Chapter 1, Properties of fluid, density, specific weight, specific gravity, viscosity (7 M) Q1(b) Reading: from Chapter 1 (8 M)

  26. Example 1.3Test 1 (Academic Session 2014/15, Sem 1) (a) Calculate the volume of mercury (S.G. = 13.54) that would have the same mass as 0.020 m3 of castor oil, which has a specific weight of 9.42 kN/m3. If the mercury has dynamic viscosity of 1.53 x 10-3Pa.s, calculate the kinematic viscosity for the mercury. (7 Marks/ Markah) (b) Explain what is meant by Newtonian and Non-Newtonian fluids. (8 Marks/ Markah) Students’ Average: Q1: 4 M; Q2: 1.8 M

  27. Example 1.3(Solution)

  28. Example 1.3(Solution)

  29. Example 1.4:Test 1 (Academic Session 2013/14, Sem 1) 1. A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass 1.56 kg. Calculate its density, specific weight and specific gravity. [6 M] 2. By using suitable example, explain what is mean by fluid viscosity. [4 M] 3. By using a simple sketch, explain what is surface tension. [5 M]

  30. Example 1.4(Solution)

  31. Example 1.4(Solution)

  32. Example 1.4(Solution)

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