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# 10 fluid dynamics

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### 10 fluid dynamics

1. 1. 1 AP Physics BAP Physics B Fluid DynamicsFluid Dynamics
2. 2. 2 College Board ObjectivesCollege Board Objectives II. FLUID MECHANICS AND THERMAL PHYSICS A. Fluid Mechanics 1. Hydrostatic pressure Students should understand the concept of pressure as it applies to fluids, so they can: a) Apply the relationship between pressure, force, and area. b) Apply the principle that a fluid exerts pressure in all directions. c) Apply the principle that a fluid at rest exerts pressure perpendicular to any surface that it contacts.
3. 3. 3 d) Determine locations of equal pressure in a fluid. e) Determine the values of absolute and gauge pressure for a particular situation. f) Apply the relationship between pressure and depth in a liquid, ∆P = ρ g ∆h Fluid mechanics, cont.Fluid mechanics, cont.
4. 4. 4 Fluid Mechanics, cont.Fluid Mechanics, cont. 2. Buoyancy Students should understand the concept of buoyancy, so they can: a) Determine the forces on an object immersed partly or completely in a liquid. b) Apply Archimedes’ principle to determine buoyant forces and densities of solids and liquids.
5. 5. 5 Fluid Mechanics, cont.Fluid Mechanics, cont. 3. Fluid flow continuity Students should understand the equation of continuity so that they can apply it to fluids in motion. 4. Bernoulli’s equation Students should understand Bernoulli’s equation so that they can apply it to fluids in motion.
6. 6. 6 Unit PlanUnit Plan • Read Chapter 10.1-10 re Fluid Mechanics in Giancoli • Assignment: • Q/4,7,11 • P/5,6,11,12,14,16,25,33,35, • and 44,46,48
7. 7. 7 10.1&2 Density &10.1&2 Density & Specific GravitySpecific Gravity • The mass density ρ of a substance is the mass of the substance divided by the volume it occupies: unit: kg/m3 ρ for aluminum 2700 kg/m3 or 2.70 g/cm3 mass can be written as m = ρV and weight as mg = ρVg Specific Gravity: ρ substance / ρ water V m =ρ
8. 8. 8 Problem 10.5Problem 10.5 5. (II) A bottle has a mass of 35.00 g when empty and 98.44 g when filled with water. When filled with another fluid, the mass is 88.78 g. What is the specific gravity of this other fluid? 5. Take the ratio of the density of the fluid to that of water, noting that the same volume is used for both liquids. ( ) ( ) fluid fluid fluid fluid water waterwater 88.78 g 35.00 g 0.8477 98.44 g 35.00 g m V m SJ m V m ρ ρ − = = = = = − SGfluid
9. 9. 9 • A fluid - a substance that flows and conforms to the boundaries of its container. • A fluid could be a gas or a liquid; however on the AP Physics B exam fluids are typically liquids which are constant in density.
10. 10. 10 AnAn ideal fluidideal fluid is assumedis assumed • to be incompressible (so that its density does not change), • to flow at a steady rate, • to be nonviscous (no friction between the fluid and the container through which it is flowing), and • flows irrotationally (no swirls or eddies).
11. 11. 11 10.3 Pressure10.3 Pressure Any fluid can exert a force perpendicular to its surface on the walls of its container. The force is described in terms of the pressure it exerts, or force per unit area: Units: N/m2 or Pa (1 Pascal*) dynes/cm2 or PSI (lb/in2) 1 atm = 1.013 x 105 Pa or 15 lbs/in2 *One atmosphere is the pressure exerted on us every day by the earth’s atmosphere. A F p =
12. 12. 12 The pressure is the same in every direction in a fluid at a given depth. Pressure varies with depth.  P = F = ρAhg so P = ρgh A A
13. 13. 13 A FLUID AT REST EXERTS PRESSURE PERPENDICULAR TO ANY SURFACE THAT IT CONTACTS. THERE IS NO PARALLEL COMPONENT THAT WOULD CAUSE A FLUID AT REST TO FLOW.
14. 14. 14 PROBLEM 10-9PROBLEM 10-9 9. (I) (a) Calculate the total force of the atmosphere acting on the top of a table that measures (b) What is the total force acting upward on the underside of the table? 9. (a) The total force of the atmosphere on the table will be the air pressure times the area of the table. ( )( ) ( )5 2 5 1.013 10 N m 1.6 m 2.9 m 4.7 10 NF PA= = × = × (b) Since the atmospheric pressure is the same on the underside of the table (the height difference is minimal), the upward force of air pressure is the same as the downward force of air on the top of the table, 5 4.7 10 N× m.2.9m6.1 ×
15. 15. 15 10.4 Atmospheric Pressure and Gauge10.4 Atmospheric Pressure and Gauge PressurePressure • The pressure p1 on the surface of the water is 1 atm, or 1.013 x 105 Pa. If we go down to a depth h below the surface, the pressure becomes greater by the product of the density of the water ρ, the acceleration due to gravity g, and the depth h. Thus the pressure p2 at this depth is h h h p2 p2 p2 p1 p1 p1 ghpp ρ+= 12
16. 16. 16 In this case, p2 is called the absolute pressure -- the total static pressure at a certain depth in a fluid, including the pressure at the surface of the fluid The difference in pressure between the surface and the depth h is gauge pressure ghpp ρ=− 12 Note that the pressure at any depth does not depend of the shape of the container, only the pressure at some reference level (like the surface) and the vertical distance below that level. h h h p2 p2 p2 p1 p1 p1
17. 17. 17 14.(II) (a) What are the total force and the absolute pressure on the bottom of a swimming pool 22.0 m by 8.5 m whose uniform depth is 2.0 m? (b) What will be the pressure against the side of the pool near the bottom? (a)The absolute pressure is given by Eq. 10-3c, and the total force is the absolute pressure times the area of the bottom of the pool. ( )( )( ) ( ) ( ) ( ) 5 2 3 3 2 0 5 2 5 2 7 1.013 10 N m 1.00 10 kg m 9.80m s 2.0 m 1.21 10 N m 1.21 10 N m 22.0 m 8.5 m 2.3 10 N P P gh F PA ρ= + = × + × = × = = × = ×
18. 18. 18 (b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the bottom, 5 2 1.21 10 N mP = ×
19. 19. 19 10.5 Pascal’s Principle10.5 Pascal’s Principle • Pascal’s Principle - if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. Applications: hydraulic lift and brakes Pout = Pin And since P = F/a Fout = Fin Aout Ain Mechanical Advantage: Fout = Aout Fin Ain
20. 20. 20 Problem 10-10Problem 10-10 10.(II) In a movie, Tarzan evades his captors by hiding underwater for many minutes while breathing through a long, thin reed. Assuming the maximum pressure difference his lungs can manage and still breathe is calculate the deepest he could have been. (See page 261.) 10.The pressure difference on the lungs is the pressure change from the depth of water ( ) ( )( ) 2 3 3 2 133N m 85mm-Hg 1 mm-Hg 1.154 m 1.2 m 1.00 10 kg m 9.80m s P P g h h g ρ ρ ∆ ∆ = ∆ → ∆ = = = ≈ ×    ÷  
21. 21. 21 10-7 Buoyancy and Archimedes’ Principle This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. The buoyant force is found to be the upward force on the same volume of water:
22. 22. 22 10-7 Buoyancy and Archimedes’ Principle The net force on the object is then the difference between the buoyant force and the gravitational force.
23. 23. 23 10-7 Buoyancy and Archimedes’ Principle If the object’s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water.
24. 24. 24 10-7 Buoyancy and Archimedes’ Principle For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid.
25. 25. 25 10-7 Buoyancy and Archimedes’ Principle This principle also works in the air; this is why hot-air and helium balloons rise.
26. 26. 26 22. (I) A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock? 22. The difference in the actual mass and the apparent mass is the mass of the water displaced by the rock. The mass of the water displaced is the volume of the rock times the density of water, and the volume of the rock is the mass of the rock divided by its density. Combining these relationships yields an expression for the density of the rock. ( ) rock actual apparent water rock water rock 3 3 3 3rock rock water 9.28kg 1.00 10 kg m 2.99 10 kg m 9.28 kg 6.18 kg m m m m V m m ρ ρ ρ ρ ρ − = ∆ = = → = = × = × ∆ −
27. 27. 27 24.(II) A crane lifts the 18,000-kg steel hull of a ship out of the water. Determine (a) the tension in the crane’s cable when the hull is submerged in the water, and (b) the tension when the hull is completely out of the water. 24.(a) When the hull is submerged, both the buoyant force and the tension force act upward on the hull, and so their sum is equal to the weight of the hull. The buoyant force is the weight of the water displaced. ( )( ) buoyant hull water buoyant hull water sub hull water hull hull hull 3 3 4 2 5 5 3 3 1 1.00 10 kg m 1.8 10 kg 9.80m s 1 1.538 10 N 1.5 10 N 7.8 10 kg m T F mg m T mg F m g V g m g g m g ρ ρ ρ ρ ρ + = → = − = − = − = − × = × − = × ≈ × ×    ÷      ÷  
28. 28. 28 24. (b)When the hull is completely out of the water, the tension in the crane’s cable must be equal to the weight of the hull. ( ) ( )4 2 5 5 1.8 10 kg 9.80m s 1.764 10 N 1.8 10 NT mg= = × = × ≈ ×
29. 29. 29 34.(III) A 5.25-kg piece of wood floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink? 34.For the combination to just barely sink, the total weight of the wood and lead must be equal to the total buoyant force on the wood and the lead. ( )0.50SG = weight buoyant wood Pb wood water Pb water wood Pb water water wood Pb water water Pb wood wood Pb Pb wood 1 1 F F m g m g V g V g m m m m m m ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ = → + = + → + = + → − = − →      ÷  ÷     ( ) water wood wood Pb wood wood water Pb Pb 1 1 1 1 1 0.50 5.25kg 5.76kg 11 11 1 11.3 SG m m m SG ρ ρ ρ ρ − − − = = = = −− −        ÷  ÷  ÷              ÷ ÷  ÷     
30. 30. 30 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity If the flow of a fluid is smooth, it is called streamline or laminar flow (a). Above a certain speed, the flow becomes turbulent (b). Turbulent flow has eddies; the viscosity of the fluid is much greater when eddies are present.
31. 31. 31 We will deal with laminar flow. The mass flow rate is the mass that passes a given point per unit time. The flow rates at any two points must be equal, as long as no fluid is being added or taken away. This gives us the equation of continuity: 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity (10-4a)
32. 32. 32 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity If the density doesn’t change – typical for liquids – this simplifies to . Where the pipe is wider, the flow is slower.
33. 33. 33 10-9 Bernoulli’s Equation A fluid can also change its height. By looking at the work done as it moves, we find: This is Bernoulli’s equation. One thing it tells us is that as the speed goes up, the pressure goes down.
34. 34. 34 36. (I) A 15-cm-radius air duct is used to replenish the air of a room every 16 min. How fast does air flow in the duct? 36. We apply the equation of continuity at constant density, Eq. 10-4b. Flow rate out of duct = Flow rate into room .cm0.2 2 ( ) ( ) ( ) ( ) ( ) 2 room room duct duct duct duct 2 2 to fill to fill room room 9.2 m 5.0 m 4.5 m 3.1m s 60 s 0.15 m 16min 1 min V V A v r v v t r t π π π = = → = = =    ÷   m4.5m5.0m2.9 ××
35. 35. 35 • 39. (II) A (inside) diameter garden hose is used to fill a round swimming pool 6.1 m in diameter. How long will it take to fill the pool to a depth of 1.2 m if water issues from the hose at a speed of • 39. The volume flow rate of water from the hose, multiplied times the time of filling, must equal the volume of the pool. inch-8 5 ?sm40.0 ( ) ( ) ( ) ( ) ( ) 2 pool pool 5 2hose "hose hose 51 2 8 " 5 3.05m 1.2m 4.429 10 s 1m 0.40m s 39.37 1day 4.429 10 s 5.1 days 60 60 24s V V Av t t A v π π = → = = = × × = × ×     ÷       ÷  
36. 36. 36 40. (II) What gauge pressure in the water mains is necessary if a firehose is to spray water to a height of 15 m? 40. Apply Bernoulli’s equation with point 1 being the water main, and point 2 being the top of the spray. The velocity of the water will be zero at both points. The pressure at point 2 will be atmospheric pressure. Measure heights from the level of point 1. ( ) ( ) ( ) 2 21 1 1 1 1 2 2 22 2 3 3 2 5 2 1 atm 2 1.00 10 kg m 9.8m s 15 m 1.5 10 N m P v gy P v gy P P gy ρ ρ ρ ρ ρ + + = + + → − = = × = ×
37. 37. 37 Visit the follow website fromVisit the follow website from Boston UniversityBoston University • http://physics.bu.edu/~duffy/py105.htm – For more information about (choose from left panel) •Pressure; Fluid Statics •Fluid Dynamics •Viscosity
38. 38. 38 At the website complete theAt the website complete the following:following: 1. Read and record important1. Read and record important equations and facts.equations and facts. 2. For each equation write the2. For each equation write the quantity for each symbolquantity for each symbol 3. Write the unit for each3. Write the unit for each quantity (symbol ok)quantity (symbol ok)
39. 39. 39 Demonstrations to ViewDemonstrations to View • http://www.csupomona.edu/~physics/ol