Physics 207 Lecture 20. Chapter 15, Fluids

Transcription

1 Chapter 15, Fluids This is an actual photo of an iceberg, taken by a rig manager for Global Marine Drilling in St. Johns, Newfoundland. The water was calm and the sun was almost directly overhead so that the diver Physics 207: Lecture 20, Pg 1 Physics 207, Lecture 20, Nov. 10 Goals: Chapter 15 Understand pressure in liquids and gases Use Archimedes principle to understand buoyancy Understand the equation of continuity Use an ideal-fluid model to study fluid flow. Investigate the elastic deformation of solids and liquids Assignment HW9, Due Wednesday, Nov. 19 th Wednesday: Read all of Chapter 16 Physics 207: Lecture 20, Pg 2 Page 1

2 Fluids (Ch. 15) At ordinary temperature, matter exists in one of three states Solid - has a shape and forms a surface Liquid - has no shape but forms a surface Gas - has no shape and forms no surface What do we mean by fluids? Fluids are substances that flow. substances that take the shape of the container Atoms and molecules are free to move. No long range correlation between positions. Physics 207: Lecture 20, Pg 3 Fluids An intrinsic parameter of a fluid Density ρ = m V units : kg/m 3 = 10-3 g/cm 3 ρ(water) = x 10 3 kg/m 3 = g/cm 3 ρ(ice) = x 10 3 kg/m 3 = g/cm 3 ρ(air) = 1.29 kg/m 3 = 1.29 x 10-3 g/cm 3 ρ(hg) = 13.6 x10 3 kg/m 3 = 13.6 g/cm 3 Physics 207: Lecture 20, Pg 4 Page 2

3 Fluids Another parameter: Pressure p = F A Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as: r F = panˆ n A Physics 207: Lecture 20, Pg 5 What is the SI unit of pressure? A. Pascal B. Atmosphere C. Bernoulli D. Young E. p.s.i. Units : 1 N/m 2 = 1 Pa (Pascal) 1 bar = 10 5 Pa 1 mbar = 10 2 Pa 1 torr = Pa 1 atm = x10 5 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in 2 (=PSI) Physics 207: Lecture 20, Pg 6 Page 3

4 Pressure vs. Depth Incompressible Fluids (liquids) When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! p(y) = p 0 - y g ρ = p 0 + d g ρ Gauge pressure (subtract p 0, usually 1 atm) p 1 y 1 y 2 mg p 0 F 1 A F 2 p 2 F 2 = F 1 + m g = F 1 + ρvg F 2 /A = F 1 /A + ρvg/a p 2 = p 1 - ρg y Physics 207: Lecture 20, Pg 7 Pressure vs. Depth For a uniform fluid in an open container pressure same at a given depth independent of the container y p(y) Fluid level is the same everywhere in a connected container, assuming no surface forces Physics 207: Lecture 20, Pg 8 Page 4

5 Pressure Measurements: Barometer Invented by Torricelli A long closed tube is filled with mercury and inverted in a dish of mercury The closed end is nearly a vacuum Measures atmospheric pressure as One 1 atm = m (of Hg) Physics 207: Lecture 20, Pg 9 Exercise Pressure What happens with two fluids?? Consider a U tube containing liquids of density ρ 1 and ρ 2 as shown: ρ 2 d I ρ 1 Compare the densities of the liquids: (A) ρ 1 < ρ 2 (B) ρ 1 = ρ 2 (C) ρ 1 > ρ 2 Physics 207: Lecture 20, Pg 10 Page 5

6 Exercise Pressure What happens with two fluids?? Consider a U tube containing liquids of density ρ 1 and ρ 2 as shown: ρ 2 d I At the red arrow the pressure must be the same on either side. ρ 1 x = ρ 2 (d 1 + y) Compare the densities of the liquids: ρ 1 y (A) ρ 1 < ρ 2 (B) ρ 1 = ρ 2 (C) ρ 1 > ρ 2 Physics 207: Lecture 20, Pg 11 Archimedes Principle Suppose we weigh an object in air (1) and in water (2). W 1 W 2? How do these weights compare? W 1 < W 2 W 1 = W 2 W 1 > W 2 Buoyant force is equal to the weight of the fluid displaced Physics 207: Lecture 20, Pg 12 Page 6

7 The Golden Crown In the first century BC the Roman architect Vitruvius related a story of how Archimedes uncovered a fraud in the manufacture of a golden crown commissioned by Hiero II, the king of Syracuse. The crown (corona in Vitruvius s Latin) would have been in the form of a wreath, such as one of the three pictured from grave sites in Macedonia and the Dardanelles. Hiero would have placed such a wreath on the statue of a god or goddess. Suspecting that the goldsmith might have replaced some of the gold given to him by an equal weight of silver, Hiero asked Archimedes to determine whether the wreath was pure gold. And because the wreath was a holy object dedicated to the gods, he could not disturb the wreath in any way. (In modern terms, he was to perform nondestructive testing). Archimedes solution to the problem, as described by Vitruvius, is neatly summarized in the following excerpt from an advertisement: The solution which occurred when he stepped into his bath and caused it to overflow was to put a weight of gold equal to the crown, and known to be pure, into a bowl which was filled with water to the brim. Then the gold would be removed and the king s crown put in, in its place. An alloy of lighter silver would increase the bulk of the crown and cause the bowl to overflow. From Physics 207: Lecture 20, Pg 13 Archimedes Principle Suppose we weigh an object in air (1) and in water (2). How do these weights compare? W 1 < W 2 W 1 = W 2 W 1 > W 2 Why? Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object. W 1 W 2? Physics 207: Lecture 20, Pg 14 Page 7

8 Sink or Float? The buoyant force is equal to the weight of the liquid that is displaced. If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. FB mg y We can calculate how much of a floating object will be submerged in the liquid: Object is in equilibrium F B = mg ρ liquid g V liquid = ρ object g V object V V liquid object = ρ ρ object liquid Physics 207: Lecture 20, Pg 15 Bar Trick What happens to the water level when the ice melts? piece of rock on top of ice A. It rises B. It stays the same C. It drops Physics 207: Lecture 20, Pg 16 Page 8

9 Exercise V 1 = V 2 = V 3 = V 4 = V 5 m 1 < m 2 < m 3 < m 4 < m 5 What is the final position of each block? Physics 207: Lecture 20, Pg 17 Exercise V 1 = V 2 = V 3 = V 4 = V 5 m 1 < m 2 < m 3 < m 4 < m 5 What is the final position of each block? Not this But this Physics 207: Lecture 20, Pg 18 Page 9

10 Exercise Buoyancy A small lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown. Pb styrofoam If you turn the styrofoam + Pb upside-down, What happens? (A) It sinks (B) styrofoam Pb (C) styrofoam Pb (D) styrofoam Pb Active Figure Physics 207: Lecture 20, Pg 19 Exercise Buoyancy A small lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown. Pb styrofoam If you turn the styrofoam + Pb upside-down, What happens (assuming density of Pb > water)? (A) It sinks (B) styrofoam Pb (C) styrofoam Pb (D) styrofoam Pb Physics 207: Lecture 20, Pg 20 Page 10

11 Exercise More Buoyancy Two identical cups are filled to the same level with water. One of the two cups has plastic balls floating in it. Cup I Which cup weighs more? Cup II (A) Cup I (B) Cup II (C) the same (D) can t tell Physics 207: Lecture 20, Pg 21 Exercise More Buoyancy Two identical cups are filled to the same level with water. One of the two cups has plastic balls floating in it. Cup I Which cup weighs more? Cup II (A) Cup I (B) Cup II (C) the same (D) can t tell Physics 207: Lecture 20, Pg 22 Page 11

12 Exercise Even More Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (ρ oil < ρ ball < ρ water ) is slowly added to the container until it just covers the ball. Relative to the water level, the ball will: Hint 1: What is the buoyant force of the part in the oil as compared to the air? water oil (A) move up (B) move down (C) stay in same place Physics 207: Lecture 20, Pg 23 Exercise Even More Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (ρ oil < ρ ball < ρ water ) is slowly added to the container until it just covers the ball. Relative to the water level, the ball will: Hint 1: What is the buoyant force of the part in the oil as compared to the air? water oil (A) move up (B) move down (C) stay in same place Physics 207: Lecture 20, Pg 24 Page 12

13 Pascal s Principle So far we have discovered (using Newton s Laws): Pressure depends on depth: p = ρ g y Pascal s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. Physics 207: Lecture 20, Pg 25 Pascal s Principle in action Consider the system shown: A downward force F 1 is applied to the piston of area A 1. F1 F 2 This force is transmitted through the liquid to create an upward force F 2. Pascal s Principle says that increased pressure from F 1 (F 1 /A 1 ) is transmitted throughout the liquid. d 1 A A 2 1 d 2 F 2 > F 1 with conservation of energy Physics 207: Lecture 20, Pg 26 Page 13

14 Exercise Hydraulics: A force amplifier Aka the lever Consider the systems shown on right. In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference d i in the liquid levels. d A M If A 2 = 2 A 1, compare d A and d B A 1 A10 V 10 = V 1 = V 2 d A A 1 = d B A 2 d A A 1 = d B 2A 1 d B M d A = d B 2 A 2 A10 Physics 207: Lecture 20, Pg 27 Home Example Hydraulics: A force amplifier Aka the lever Consider the system shown on right. Blocks of mass M are placed on the piston of both cylinders. The small & large cylinders displace distances d 1 and d 2 Compare the forces on disks A 1 & A 2 ignoring the mass and weight of the fluid in this process M A 1 d 1 M A2 W 2 = - M g d 2 = - F 2 d 2 V 1 = V 2 = A 1 d 1 = A 2 d 2 W 1 = M g d 1 = F 1 d 1 W 1 + W 2 = 0 = F 1 d 1 - F 2 d 2 F 2 d 2 = F 1 d 1 F 2 = F 1 d 1 / d 2 = F 1 A 2 / A 1 d 1 / d 2 = A 2 / A 1 implying P 1 = P 2 = F 1 /A 1 = F 2 /A 2 Physics 207: Lecture 20, Pg 28 Page 14

15 Fluids in Motion To describe fluid motion, we need something that describes flow: Velocity v There are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / ideal Physics 207: Lecture 20, Pg 29 Types of Fluid Flow Laminar flow Each particle of the fluid follows a smooth path The paths of the different particles never cross each other The path taken by the particles is called a streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed Physics 207: Lecture 20, Pg 30 Page 15

16 Types of Fluid Flow Laminar flow Each particle of the fluid follows a smooth path The paths of the different particles never cross each other The path taken by the particles is called a streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed Physics 207: Lecture 20, Pg 31 Onset of Turbulent Flow The SeaWifS satellite image of a von Karman vortex around Guadalupe Island, August 20, 1999 Physics 207: Lecture 20, Pg 32 Page 16

17 Ideal Fluids Fluid dynamics is very complicated in general (turbulence, vortices, etc.) Consider the simplest case first: the Ideal Fluid No viscosity - no flow resistance (no internal friction) Incompressible - density constant in space and time Simplest situation: consider ideal fluid moving with steady flow - velocity at each point in the flow is constant in time streamline A 2 A 1 In this case, fluid moves on streamlines v 1 v 2 Physics 207: Lecture 20, Pg 33 Ideal Fluids Streamlines do not meet or cross Velocity vector is tangent to streamline streamline A 2 A 1 Volume of fluid follows a tube of flow bounded by streamlines v 1 v 2 Streamline density is proportional to velocity Flow obeys continuity equation Volume flow rate Q = A v is constant along flow tube. A 1 v 1 = A 2 v 2 Follows from mass conservation if flow is incompressible. Physics 207: Lecture 20, Pg 34 Page 17

18 Exercise Continuity A housing contractor saves some money by reducing the size of a pipe from 1 diameter to 1/2 diameter at some point in your house. v 1 v 1/2 Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1 diameter pipe, how fast is the water going in the 1/2 pipe? (A) 2 v 1 (B) 4 v 1 (C) 1/2 v 1 (D) 1/4 v 1 Physics 207: Lecture 20, Pg 35 Exercise Continuity v 1 v 1/2 (A) 2 v 1 (B) 4 v 1 (C) 1/2 v 1 (D) 1/4 v 1 For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast. But if the water is moving four times as fast then it has 16 times as much kinetic energy. Something must be doing work on the water (the pressure drops at the neck and we recast the work as P V = (F/A) (A x) = F x ) Physics 207: Lecture 20, Pg 36 Page 18

19 Lecture 20, Nov. 10 Question to ponder: Does heavy water (D 2 0) ice sink or float? Assignment HW9, Due Wednesday, Nov. 19 th Wednesday: Read all of Chapter 16 Next slides are possible for Wednesday Physics 207: Lecture 20, Pg 37 Conservation of Energy for Ideal Fluid Recall the standard work-energy relation W = K = K f K i Apply the principle to a section of flowing fluid with volume V and mass m = ρ V (here W is work done on fluid) Net work by pressure difference over x ( x 1 = v 1 t) W = F 1 x 1 F 2 x 2 = (F 1 /A 1 ) (A 1 x 1 ) (F 2 /A 2 ) (A 2 x 2 ) = P 1 V 1 P 2 V 2 v 2 and V 1 = V 2 = V (incompressible) y 2 W = (P 1 P 2 ) V v 1 V p 2 y 1 p 1 Physics 207: Lecture 20, Pg 38 Page 19

20 W Conservation of Energy for Ideal Fluid = (P 1 P 2 ) V and W = ½ m v 2 2 ½ m v 1 2 = ½ (ρ V) v 2 2 ½ (ρ V) v 1 2 (P 1 P 2 ) = ½ ρ v 22 ½ ρ v 1 2 v 2 P 1 + ½ ρ v 1 2 = P 2 + ½ ρ v 2 2 = const. y 2 v 1 V p 2 y 1 p 1 Bernoulli Equation P 1 + ½ ρ v ρ g y 1 = constant Physics 207: Lecture 20, Pg 39 This leads to Conservation of Energy for Ideal Fluid P 1 + ½ ρ v 12 = P 2 + ½ ρ v 22 = const. and with height variations v 2 y 2 v 1 V p 2 y 1 p 1 Bernoulli s Equation P 1 + ½ ρ v ρ g y 1 = constant Physics 207: Lecture 20, Pg 40 Page 20

21 Bernoulli s Principle A housing contractor saves some money by reducing the size of a pipe from 1 diameter to 1/2 diameter at some point in your house. v 1 v 1/2 2) What is the pressure in the 1/2 pipe relative to the 1 pipe? (A) smaller (B) same (C) larger Physics 207: Lecture 20, Pg 41 Cavitation Venturi result In the vicinity of high velocity fluids, the pressure can gets so low that the fluid vaporizes. Physics 207: Lecture 20, Pg 42 Page 21

22 Applications of Fluid Dynamics Streamline flow around a moving airplane wing Lift is the upward force on the wing from the air Drag is the resistance The lift depends on the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal higher velocity lower pressure lower velocity higher pressure Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid. Physics 207: Lecture 20, Pg 43 Elastic properties of solids : Some definitions Young s modulus: measures the resistance of a solid to a change in its length. L 0 L F elasticity in length Y = tensile tensile stress strain = F / A0 L / L 0 volume elasticity Bulk modulus: measures the resistance of solids or liquids to changes in their volume. F / A0 B = V / V 0 V 0 Physics 207: Lecture 20, Pg 44 F V 0 - V Page 22

23 Physics 207: Lecture 20, Pg 45 EXAMPLE An irrigation system QUESTION: Physics 207: Lecture 20, Pg 46 Page 23

24 EXAMPLE An irrigation system Physics 207: Lecture 20, Pg 47 EXAMPLE An irrigation system Physics 207: Lecture 20, Pg 48 Page 24

25 EXAMPLE An irrigation system Physics 207: Lecture 20, Pg 49 EXAMPLE An irrigation system Physics 207: Lecture 20, Pg 50 Page 25

26 Elasticity Physics 207: Lecture 20, Pg 51 Elasticity F/A is proportional to L/L. We can write the proportionality as The proportionality constant Y is called Young s modulus. The quantity F/A is called the tensile stress. The quantity L/L, the fractional increase in length, is called strain. With these definitions, we can write Physics 207: Lecture 20, Pg 52 Page 26

27 EXAMPLE Stretching a wire QUESTIONS: Physics 207: Lecture 20, Pg 53 EXAMPLE Stretching a wire Physics 207: Lecture 20, Pg 54 Page 27

28 EXAMPLE Stretching a wire Physics 207: Lecture 20, Pg 55 Volume Stress and the Bulk Modulus Physics 207: Lecture 20, Pg 56 Page 28

29 Volume Stress and the Bulk Modulus A volume stress applied to an object compresses its volume slightly. The volume strain is defined as V/V, and is negative when the volume decreases. Volume stress is the same as the pressure. where B is called the bulk modulus. The negative sign in the equation ensures that the pressure is a positive number. Physics 207: Lecture 20, Pg 57 The figure shows volume flow rates (in cm 3 /s) for all but one tube. What is the volume flow rate through the unmarked tube? Is the flow direction in or out? A. 1 cm 3 /s, in B. 1 cm 3 /s, out C. 10 cm 3 /s, in D. 10 cm 3 /s, out E. It depends on the relative size of the tubes. Physics 207: Lecture 20, Pg 58 Page 29

30 Rank in order, from highest to lowest, the liquid heights h 1 to h 4 in tubes 1 to 4. The air flow is from left to right. A. h 1 > h 2 = h 3 = h 4 B. h 2 > h 4 > h 3 > h 1 C. h 2 = h 3 = h 4 > h 1 D. h 3 > h 4 > h 2 > h 1 E. h 1 > h 3 > h 4 > h 2 Physics 207: Lecture 20, Pg 59 Page 30