Fluids, Thermodynamics, Waves, and Optics Fluids Lana Sheridan De Anza College April 10, 2018
Overview static fluids pressure liquid pressure Pascal s law
Elastic Properties of Solids We are considering 3 kinds of elastic modulus in terms of: Stress the external force acting on an object per unit cross-sectional area F /A Strain the fractional amount of change in shape of the object or material For each: elastic modulus = stress strain 1 See Serway & Jewett, Chapter 12 section 4.
Shear Modulus: Shape Elasticity Shear modulus, S, is a measure of the resistance to motion of parallel layers within a solid. Y = F /A x/h where F /A is the shear stress: F is a tangential force, parallel to the surface F os applied to. x/h is the shear strain: x is the distance the face is displaced and h is the perpendicular distance between faces. h x A S F S F Fixed face
Fluids Before looking at the third kind of elastic modulus, we will introduce fluids. The previous two elastic moduli apply to solids. The bulk modulus is relevant for both solids and fluids.
Fluids The term fluid encompasses both liquids and gases. It is a collection of molecules that are only weakly influenced by intermolecular forces and thus can flow over each other. 1 Figure from Wikipedia, user Duk.
Fluids The term fluid encompasses both liquids and gases. It is a collection of molecules that are only weakly influenced by intermolecular forces and thus can flow over each other. Formally, Fluids deform continuously under shearing stress; have a shear modulus of zero. 1 Figure from Wikipedia, user Duk.
r substance to change t the substance as a Fluid Statics held together We firstby consider fluid statics: fluids at rest in a container. r. Both liquids and ples and analysis ics of a fluid at rest, mics. At any point on the surface of the object, the force exerted by the fluid is perpendicular to the surface of the object. s those discussed in object submerged in ides. In other words, Figure 14.1 The forces exerted cular Fluids to the surfaces will exert pressure by a fluid on on objects the surfaces submerged of a sub-imerged them and also in Section the12.4. walls of the container. object. 417
held together by r. Both Pressure liquids and the object, the force exerted by the fluid is perpendicular to the surface of the object. ples and analysis ics of a fluid at rest, mics. s those discussed in object submerged in ides. In other words, Figure 14.1 The forces exerted cular to the surfaces by a fluid on the surfaces of a submerged object. in Section Pressure 12.4. is the normal force per unit area on a surface: P = F A 417
Pressure It is possible that the pressure on a surface varies over the surface. Then we will need to extend our definition: δf = P δa δf is the force on a tiny area δa. As δa 0, the pressure can be a continuous function of the position on the surface.
Pressure It is possible that the pressure on a surface varies over the surface. Then we will need to extend our definition: δf = P δa δf is the force on a tiny area δa. As δa 0, the pressure can be a continuous function of the position on the surface. Units: the Pascal (after Blaise Pascal), symbol Pa 1 Pa = 1 N/m 2
Pressure Pressure is a scalar quantity. We relate it to force, a vector, by: δf = P δa ˆn where ˆn is a unit vector perpendicular to the small area δa. In a gas or liquid, its underlying cause is molecular collisions: 1 Diagram by Brant Carlson, on Wikipedia.
Question Quick Quiz 14.1 1 Suppose you are standing directly behind someone who steps back and accidentally stomps on your foot with the heel of one shoe. Would you be better off if that person were A a large, male professional basketball player wearing sneakers or B a petite woman wearing spike-heeled shoes? 1 Page 418, Serway & Jewett
Question Quick Quiz 14.1 1 Suppose you are standing directly behind someone who steps back and accidentally stomps on your foot with the heel of one shoe. Would you be better off if that person were A a large, male professional basketball player wearing sneakers or B a petite woman wearing spike-heeled shoes? 1 Page 418, Serway & Jewett
Ambient Atmospheric Pressure The air around us exerts force on us, the walls of the room, the floor, etc. Is it a large pressure?
Ambient Atmospheric Pressure The air around us exerts force on us, the walls of the room, the floor, etc. Is it a large pressure? In a sense, yes! P 0 = P atm = 1.013 10 5 Pa
Bulk Modulus: Volume Elasticity astic Properties of Solids 375 en an rgoes ratio e surtail in object lume actered as (12.8) mber. uses a s in a recips. No use a e or a - Bulk modulus, S, is a measure of the resistance to compression of a material. B = P V /V i The negative sign ensures B will be a positive number. Bulk modulus S F left S F front S F top S F bottom V i S F back S F right V i V The cube undergoes a change in where F /A is the volume stress or pressure change. V /V i is the volume strain: V is the change from the initial volume V i. The reciprocal of the bulk modulus, 1/B, is the compressibility of the material.
Pressure in a Liquid in a Gravitational Field In a uniform gravitational field, liquid pressure depends on depth. P liq = ρgh where ρ = m/v is the mass density of the liquid and h is the depth. It does not depend on the total amount of water involved, just the depth of water.
Liquid Pressure A slice of liquid of cross section A at a depth h must support all the water in a column directly above it. The force exerted downward by the column of water is F = mg = ρvg.
Liquid Pressure F = mg = ρvg = ρahg 1 Figure from HyperPhysics.
Liquid Pressure F = mg = ρvg = ρahg Pressure, P liq = F A = ρahg A = ρgh. 1 Figure from HyperPhysics.
Total Pressure The liquid pressure only expresses the pressure due to the weight of the fluid above. However, this is not the total pressure in most circumstances, eg. diving on earth.
Total Pressure The liquid pressure only expresses the pressure due to the weight of the fluid above. However, this is not the total pressure in most circumstances, eg. diving on earth. The total pressure or absolute pressure is the sum of the pressure due to the liquid and the pressure due to the atmosphere. P total = P 0 + ρgh where P 0 = P atm = 1.013 10 5 Pa.
Pressure varies with Depth
Pascal s Barrel
Density of Water For water: ρ w = 1.00 10 3 kg/m 3 That is ρ w = 1 g/cm 3.
Density of Water For water: ρ w = 1.00 10 3 kg/m 3 That is ρ w = 1 g/cm 3. Originally, the gram was defined to be the mass of one cubic centimeter of water at the melting point of water.
Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 2 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.
Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 2 Density of water: ρ w = 1000 kg/m 3 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.
Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 2 Density of water: ρ w = 1000 kg/m 3 P liq = 2.16 10 6 Pa P total 2.3 10 6 Pa 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.
Questions Calculate the water pressure at the base of the Hoover Dam. The depth of water behind the dam is 220 m. 2 Density of water: ρ w = 1000 kg/m 3 P liq = 2.16 10 6 Pa P total 2.3 10 6 Pa Now consider, if the dam is 380 m long, what is the total force exerted by the water on the dam? 3 2 Question from Hewitt, Conceptual Physics, 11th ed. 3 See example 14.4, page 422, Serway & Jewett, 9th ed.
he middle ear. Using this technique equalizes the pressure. Questions Now consider, if the dam is 380 m long, what is the total force exerted by the water on the dam? 14.5). Determine the h y cannot calculate the pressure in the water e dam also increases., we must use integraproblem. H ttom of the dam. We x a distance y above the O on each such strip is Figure 14.5 3 (Example 14.4) Water See example 14.4, page 422, Serway & Jewett, 9th ed. w dy y
he middle ear. Using this technique equalizes the pressure. Questions Now consider, if the dam is 380 m long, what is the total force exerted by the water on the dam? 14.5). Determine the h y cannot calculate the pressure in the water e dam also increases., we must use integraproblem. H ttom of the dam. We x a distance y above the O on each such strip is Figure 14.5 3 (Example 14.4) Water See example 14.4, page 422, Serway & Jewett, 9th ed. w dy y F = 9.0 10 10 N
Pressure in a liquid We have this expression for total pressure: P total = P 0 + ρgh What if the pressure at the surface of the liquid, P 0, was increased to P 1. How would we expect this relation to change?
Pressure in a liquid We have this expression for total pressure: P total = P 0 + ρgh What if the pressure at the surface of the liquid, P 0, was increased to P 1. How would we expect this relation to change? P total = P 1 + ρgh The differences in pressure between the different layers of liquid remain the same, but the pressure at each depth h increases.
Pascal s Law This simple idea is captured by Pascal s Law. Pascal s law applied to confined, incompressible fluids. Pascal s Law A change in pressure applied to one part of an (incompressible) fluid is transmitted undiminished to every point of the fluid.
Pascal s Law This simple idea is captured by Pascal s Law. Pascal s law applied to confined, incompressible fluids. Pascal s Law A change in pressure applied to one part of an (incompressible) fluid is transmitted undiminished to every point of the fluid. This does not mean that the pressure is the same at every point in the fluid. It means that if the pressure is increased at one point in the fluid, it increases by the same amount at all other points.
Pascal s Principle Since the changes in pressures at the left end and the right end are the same: Since A 2 > A 1, F 2 > F 1. P 1 = P 2 F 1 = F 2 A 1 A 2
Hydraulic Lift This has applications: 1 Figure from hyperphysics.phy-astr.gsu.edu.
Question If a pair of pistons are connected on either end of a hydraulic tube. The first has area 0.2 m 2 and the second has an area of 4 m 2. A force of 30 N is applied the first piston. What is the force exerted by the second piston on a mass that rests on it?
Question If a pair of pistons are connected on either end of a hydraulic tube. The first has area 0.2 m 2 and the second has an area of 4 m 2. A force of 30 N is applied the first piston. What is the force exerted by the second piston on a mass that rests on it? If the first piston is depressed a distance of 1 m by the 30 N force, how far does the second piston rise?
Summary static fluids pressure Pascal s law Test Tuesday, April 17, in class. Collected Homework due Monday, April 16. (Uncollected) Homework Ch 12, onward from page 382, Obj Ques: 9; Probs: 27, 29. Ch 14, onward from page 435, OQs: 1; CQs: 1, 3, 7; Probs: 1, 3, 7, 8, 9, 11, 15, 18