A Physical Introduction to Fluid Mechanics

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A Physical Introduction to Fluid Mechanics A Physical Introduction to Fluid Mechanics Study Guide and Practice Problems Spring 2014 A Physical Introduction to Fluid Mechanics Study Guide and Practice Problems Spring 2014 by Alexander J. Smits Professor of Mechanical and Aerospace Engineering Princeton University and Professorial Fellow Department of Mechanical and Aerospace Engineering Monash University December 19, 2013 Copyright A.J. Smits c 2014 Contents 1 Introduction 1 1.1 Study Guide . .1 1.2 Worked Examples . .1 Problems . .9 2 Fluid Statics 13 2.1 Study Guide . 13 2.2 Worked Examples . 14 Problems . 22 3 Equations of Motion in Integral Form 47 3.1 Study Guide . 47 3.2 Worked Examples . 47 Problems . 60 4 Kinematics and Bernoulli's Equation 81 4.1 Study Guide . 81 4.2 Worked Examples . 81 Problems . 84 5 Differential Equations of Motion 99 5.1 Study Guide . 99 5.2 Worked Examples . 99 Problems . 104 6 Irrotational, Incompressible Flows 111 6.1 Study Guide . 111 6.2 Worked Examples . 111 Problems . 113 7 Dimensional Analysis 117 7.1 Study Guide . 117 7.2 Worked Examples . 117 Problems . 122 8 Viscous Internal Flows 133 8.1 Study Guide . 133 8.2 Worked Examples . 133 Problems . 139 v vi CONTENTS 9 Viscous External Flows 151 9.1 Study Guide . 151 9.2 Worked Examples . 152 Problems . 154 10 Open Channel Flow 163 10.1 Study Guide . 163 10.2 Worked Examples . 164 Problems . 167 11 Compressible Flow 185 11.1 Study Guide . 185 11.2 Worked Examples . 185 Problems . 190 12 Turbomachines 193 12.1 Worked Examples . 193 Problems . 197 13 Environmental Fluid Mechanics 201 Problems . 201 Answers to Selected Problems 203 Chapter 1 Introduction 1.1 Study Guide • A fluid is defined as a material that deforms continuously and permanently under the application of a shearing stress. • The pressure at a point in a fluid is independent of the orientation of the surface passing through the point; the pressure is isotropic. • The force due to a pressure p acting on one side of a small element of surface dA defined by a unit normal vector n is given by −pndA. • Pressure is transmitted through a fluid at the speed of sound. • The units we use depend on whatever system we have chosen, and they include quan- tities like feet, seconds, newtons and pascals. In contrast, a dimension is a more abstract notion, and it is the term used to describe concepts such as mass, length and time. • The specific gravity (SG) of a solid or liquid is the ratio of its density to that of water at the same temperature. • A Newtonian fluid is one where the viscous stress is proportional to the rate of strain (velocity gradient). The constant of proportionality is the viscosity, µ, which is a property of the fluid, and depends on temperature. • At the boundary between a solid and a fluid, the fluid and solid velocities are equal; this is called the \no-slip condition." As a consequence, for large Reynolds numbers (>> 1), boundary layers form close to the solid boundary. In the boundary layer, large velocity gradients are found, and so viscous effects are important. • At the interface between two fluids, surface tension may become important. Surface tension leads to the formation of a meniscus, drops and bubbles, and the capillary rise observed in small tubes, because surface tension can resist pressure differences across the interface. 1.2 Worked Examples Example 1.1: Units, and converting between units Consider a rectangular block with dimensions 300 mm × 100 mm × 25 mm, of mass 10 kg, resting on a surface (Figure 1.1). The pressure acting over the area of contact can be found 1 2 CHAPTER 1. INTRODUCTION as follows. Solution: Since pressure is a stress, it has dimensions of force per unit area. When in position (a), the force exerted on the table is equal to the weight of the block (= mass × gravitational acceleration = 98 N), and the average pressure over the surface in contact with the table is given by 98 N=m2 = 39; 200 P a 100 × 25 × 10−6 In position (b), the force exerted on the table is still equal to 98 N, but the average pres- sure over the surface in contact with the table is reduced to 98300 × 25 × 10−6 N=m2, that is, 13; 067 P a. We can repeat this example using engineering units. Let's take a rectangular block, made of a different material with dimensions 12 in: × 4 in: × 1 in:, of mass 20 lbm (this is similar to the case shown in Figure 1.1). Find the pressure acting over the area in the BG system. Solution: The unit lbm is not part of the engineering system of units (see Table 1.1), so we first convert it to slugs, where mass in lb 20 mass in slugs = m = slug = 0:622 slug 32:2 32:2 So 20 lbm = 0:622 slug: When in position (a), the force exerted on the table is equal to the weight of the block (= mass × gravitational acceleration = 20 lbf ), and the average pressure over the surface 2 in contact with the table is 20/(4 × 1) lbf =in: , that is, 5 psi. In position (b), the force exerted on the table is still equal to 20 lbf , but now the average pressure over the surface 2 in contact with the table is 20/(12 × 1) lbf =in: , that is, 1:67 psi. Example 1.2: Hydraulic presses and hoists A hydraulic press uses the transmissibility of fluid pressure to produce large forces. A simple press consists of two connected cylinders of significantly different size, each fitted with a piston and filled with either oil or water (see Figure 1.2). The pressures produced by hydraulic devices are typically hundreds or thousands of psi, so that the weight of the fluid can usually be neglected. If there is an unobstructed passage connecting the two cylinders, then p1 ≈ p2: Since pressure = (magnitude of force)/area, F f A = ; so that F = f A a a Figure 1.1: Pressure exerted by a weight resting on a surface. All dimensions in millimeters. 1.2. WORKED EXAMPLES 3 Figure 1.2: Hydraulic press. Figure 1.3: Hydraulic drum brake system. The pressure transmission amplifies the applied force; a hydraulic press is simply a hydraulic lever. A hydraulic hoist is basically a hydraulic press that is turned around. In a typical garage hoist, compressed air is used (instead of an actuating piston) to force oil through the connecting pipe into the cylinder under a large piston, which supports the car. A lock-valve is usually placed in the connecting pipe, and when the hoist is at the right height the valve is closed, holding the pressure under the cylinder and maintaining the hoist at a constant height. A similar application of the transmissibility of pressure is used in hydraulic brake systems. Here, the force is applied by a foot pedal, increasing the pressure in a \master" cylinder, which in turn transmits the pressure to each brake or \slave" cylinder (Figure 1.3). A brake cylinder has two opposing pistons, so that when the pressure inside the cylinder rises the two pistons move in opposite directions. In a drum brake system, each brake shoe is pivoted at one end, and attached to one of the pistons of the slave cylinder at the other end. As the piston moves out, it forces the brake shoe into contact with the brake drum. Similarly, in a disk brake system, there are two brake pads, one on each side of the disk, and the brake cylinder pushes the two brake pads into contact with the disk. Example 1.3: Finding the pressure in a fluid Consider the piston and cylinder illustrated in Figure 1.4. If the piston had a mass m = 1 kg, and an area A = 0:01 m2, what is the pressure p of the gas in the cylinder? Atmospheric pressure pa acts on the outside of the container. 4 CHAPTER 1. INTRODUCTION Solution: The piston is not moving so that it is in equilibrium under the force due to its own weight, the force due to the gas pressure inside of the piston (acting up), and the force due to air pressure on the outside of the piston (acting down). The weight of the piston = mg, where g is the acceleration due to gravity (9:8 m=s2, or 32:1 ft=s2). The force due to pressure = pressure × area of piston. Therefore 2 pA − paA = mg = 1 kg × 9:8 m =s = 9:8 N That is, 9:8 N p − p = = 980 P a a 0:01 m2 where P a = pascal = N=m2. What is this excess pressure in psi (pounds per square inch)? A standard atmosphere 2 has a pressure of 14:7 lbf =in: , or 101; 325 P a (see Table 2.1). Therefore, 980 P a is equal to 14:7 × 980=101325 psi = 0:142 psi. Example 1.4: Force due to pressure In the center of a hurricane, the pressure can be very low. Find the force acting on the wall of a house, measuring 10 ft by 20 ft, when the pressure inside the house is 30:0 in of mercury, and the pressure outside is 26:3 in: of mercury. Express the answer in lbf and N. Solution: A mercury barometer measures the local atmospheric pressure. A standard 2 2 atmosphere has a pressure of 14:7 lbf =in: , or 101; 325 P a or N=m Table 2.1).
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