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Chapter 9 Fluids.

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Presentation on theme: "Chapter 9 Fluids."— Presentation transcript:

1 Chapter 9 Fluids

2 Ch09b - Fluids-Revised: 3/6/2010
Chapter 9: Fluids Introduction to Fluids Pressure Measurement of Pressure Pascal’s Principle Gravity and Fluid Pressure Archimedes’ Principle Continuity Equation Bernoulli’s Equation Viscosity and Viscous Drag Surface Tension MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

3 Ch09b - Fluids-Revised: 3/6/2010
Pressure Pressure arises from the collisions between the particles of a fluid with another object (container walls for example). There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

4 Ch09b - Fluids-Revised: 3/6/2010
By Newton’s 3rd Law, there is a force on the wall due to the particle. Pressure is defined as The units of pressure are N/m2 and are called Pascals (Pa). Note: 1 atmosphere (atm) = kPa MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

5 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.1): Someone steps on your toe, exerting a force of 500 N on an area of 1.0 cm2. What is the average pressure on that area in atmospheres? MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Measuring Pressure A manometer is a U-shaped tube that is partially filled with liquid. Both ends of the tube are open to the atmosphere. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

7 Ch09b - Fluids-Revised: 3/6/2010
A container of gas is connected to one end of the U-tube If there is a pressure difference between the gas and the atmosphere, a force will be exerted on the fluid in the U-tube. This changes the equilibrium position of the fluid in the tube. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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From the figure: At point C Also The pressure at point B is the pressure of the gas. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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A Barometer The atmosphere pushes on the container of mercury which forces mercury up the closed, inverted tube. The distance d is called the barometric pressure. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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From the figure and Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Pascal’s Principle A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid. (This is useful in making a hydraulic lift.) MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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The applied force is transmitted to the piston of cross-sectional area A2 here. Apply a force F1 here to a piston of cross-sectional area A1. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Mathematically, MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A2/A1) are 1, 10, and 100. Using Pascal’s Principle: 50,000 N 100 5000 N 10 500 N 1 F2 MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

15 Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder A cylinder of fluid x y P1A Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 9, Questions 4 and 5. w P2A MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

16 Ch09b - Fluids-Revised: 3/6/2010
Apply Newton’s 2nd Law to the fluid cylinder: If P1 (the pressure at the top of the cylinder) is known, then the above expression can be used to find the variation of pressure with depth in a fluid. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

17 Ch09b - Fluids-Revised: 3/6/2010
If the top of the fluid column is placed at the surface of the fluid, then P1 = Patm if the container is open. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

18 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.19): At the surface of a freshwater lake, the pressure is 105 kPa. (a) What is the pressure increase in going 35.0 m below the surface? Part b of this problem is not included. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

19 Ch09b - Fluids-Revised: 3/6/2010
Example: The surface pressure on the planet Venus is 95 atm. How far below the surface of the ocean on Earth do you need to be to experience the same pressure? The density of seawater is 1025 kg/m3. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

20 Archimedes’ Principle
w F2 F1 x y An FBD for an object floating submerged in a fluid. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 9, Questions 6, 7, 14, 18, and 20. The total force on the block due to the fluid is called the buoyant force. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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The magnitude of the buoyant force is: From before: The result is MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Archimedes’ Principle: A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

23 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.28): A flat-bottomed barge loaded with coal has a mass of 3.0105 kg. The barge is 20.0 m long and 10.0 m wide. It floats in fresh water. What is the depth of the barge below the waterline? x y w FB Apply Newton’s 2nd Law to the barge: FBD for the barge Symbols defined: mw is the mass of water displaced, Vw is the volume of water displaced by the barge, w is the density of fresh water, and mb is the mass of the barge. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

24 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.40): A piece of metal is released under water. The volume of the metal is 50.0 cm3 and its specific gravity is 5.0. What is its initial acceleration? (Note: when v = 0, there is no drag force.) x y w FB Apply Newton’s 2nd Law to the piece of metal: FBD for the metal The magnitude of the buoyant force equals the weight of the fluid displaced by the metal. Solve for a: MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Example continued: Since the object is completely submerged V=Vobject. where water = 1000 kg/m3 is the density of water at 4 °C. Given MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Fluid Flow A moving fluid will exert forces parallel to the surface over which it moves, unlike a static fluid. This gives rise to a viscous force that impedes the forward motion of the fluid. A steady flow is one where the velocity at a given point in a fluid is constant. V1 = constant V2 = constant v1v2 Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 9, Questions 8 and 11. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Steady flow is laminar; the fluid flows in layers. The path that the fluid in these layers takes is called a streamline. Streamlines do not cross. An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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The continuity equation—Conservation of mass. The amount of mass that flows though the cross-sectional area A1 is the same as the mass that flows through cross-sectional area A2. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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is the mass flow rate (units kg/s) is the volume flow rate (units m3/s) The continuity equation is If the fluid is incompressible, then 1= 2. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Example (text problem 9.41): A garden hose of inner radius 1.0 cm carries water at 2.0 m/s. The nozzle at the end has radius 0.20 cm. How fast does the water move through the constriction? MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

31 Ch09b - Fluids-Revised: 3/6/2010
Bernoulli’s Equation Bernoulli’s equation is a statement of energy conservation. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 9, Questions 9, 10, and 19. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Work per unit volume done by the fluid Potential energy per unit volume Points 1 and 2 must be on the same streamline Kinetic energy per unit volume MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

33 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.49): A nozzle is connected to a horizontal hose. The nozzle shoots out water moving at 25 m/s. What is the gauge pressure of the water in the hose? Neglect viscosity and assume that the diameter of the nozzle is much smaller than the inner diameter of the hose. Let point 1 be inside the hose and point 2 be outside the nozzle. The hose is horizontal so y1 = y2. Also P2 = Patm. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

34 Ch09b - Fluids-Revised: 3/6/2010
Example continued: Substituting: v2 = 25 m/s and v1 is unknown. Use the continuity equation. Since d2<<d1 it is true that v1<<v2. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Example continued: MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Viscosity A real fluid has viscosity (fluid friction). This implies a pressure difference needs to be maintained across the ends of a pipe for fluid to flow. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

37 Ch09b - Fluids-Revised: 3/6/2010
Viscosity also causes the existence of a velocity gradient across a pipe. A fluid flows more rapidly in the center of the pipe and more slowly closer to the walls of the pipe. The volume flow rate for laminar flow of a viscous fluid is given by Poiseuille’s Law. where  is the viscosity MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

38 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.55): A hypodermic syringe attached to a needle has an internal radius of mm and a length of 3.00 cm. The needle is filled with a solution of viscosity 2.0010-3 Pa sec; it is injected into a vein at a gauge pressure of 16.0 mm Hg. Neglect the extra pressure required to accelerate the fluid from the syringe into the entrance needle. (a) What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of mL/sec? Solve Poiseuille’s Law for the pressure difference: MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

39 Ch09b - Fluids-Revised: 3/6/2010
Example continued: This pressure difference is between the fluid in the syringe and the fluid in the vein. The pressure in the syringe is Ps is the pressure in the syringe and Pv is the pressure in the vein. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

40 Ch09b - Fluids-Revised: 3/6/2010
Example continued: (b) What force must be applied to the plunger, which has an area of 1.00 cm2? The result of (a) gives the force per unit area on the plunger so the force is just F = PA = N. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

41 Ch09b - Fluids-Revised: 3/6/2010
Viscous Drag The viscous drag force on a sphere is given by Stokes’ law. Where  is the viscosity of the fluid, r is the radius of the sphere, and v is the velocity of the sphere. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

42 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.62): A sphere of radius 1.0 cm is dropped into a glass cylinder filled with a viscous liquid. The mass of the sphere is 12.0 g and the density of the liquid is 1200 kg/m3. The sphere reaches a terminal speed of 0.15 m/s. What is the viscosity of the liquid? FD w x y FB Apply Newton’s Second Law to the sphere FBD for sphere FD is the drag force as given by Stokes’ law and FB is the buoyant force as given by Archimedes principle. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Example continued: When v = vterminal, a = 0 and Solving for  Symbol definitions: vt is the terminal velocity, ml is the mass of liquid displaced by the sphere, Vl is the volume of liquid displaced by the sphere and is equal to the volume of the sphere Vs, ms is the mass of the sphere, and l is the density of the liquid. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

44 Ch09b - Fluids-Revised: 3/6/2010
Surface Tension The surface of a fluid acts like a a stretched membrane (imagine standing on a trampoline). There is a force along the surface of the fluid. The surface tension is a force per unit length. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 9, Question 16. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

45 Ch09b - Fluids-Revised: 3/6/2010
Example (text problem 9.70): Assume a water strider has a roughly circular foot of radius 0.02 mm. The water strider has 6 legs. (a) What is the maximum possible upward force on the foot due to the surface tension of the water? The water strider will be able to walk on water if the net upward force exerted by the water equals the weight of the insect. The upward force is supplied by the water’s surface tension. From the text =0.070 N/m. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

46 Ch09b - Fluids-Revised: 3/6/2010
Example continued: (b) What is the maximum mass of this water strider so that it can keep from breaking through the water surface? To be in equilibrium, each leg must support one-sixth the weight of the insect. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

47 Ch09b - Fluids-Revised: 3/6/2010
Summary Pressure and its Variation with Depth Pascal’s Principle Archimedes Principle Continuity Equation (conservation of mass) Bernoulli’s Equation (conservation of energy) Viscosity and Viscous Drag Surface Tension MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

48 Quick Questions

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Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged? 1. Rise 2. Fall 3. Remain unchanged MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

50 1. Rise 2. Fall 3. Remain unchanged
Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged? 1. Rise 2. Fall 3. Remain unchanged MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball 1. rises. 2. falls. 3. remains in place. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be 1. half. 2. the same. 3. twice. 4. more than twice. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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A pair of identical balloons are inflated with air and suspended on the ends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is 1. upset and the stick rotates clockwise. 2. upset and the stick rotates counter-clockwise. 3. unchanged. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

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Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water. Now if you push it beneath the surface, it will 1. sink. 2. return to the surface. 3. stay at the depth to which it is pushed. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

59 Ch09b - Fluids-Revised: 3/6/2010
MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

60 Extras

61 The density of the block of wood floating in water is
1. greater than the density of water. 2. equal to the density of water. 3. less than half that of water. 4. more than half the density of water. 5. … not enough information is given. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

62 Ch09b - Fluids-Revised: 3/6/2010
The density of the block of wood floating in water is 1. greater than the density of water. 2. equal to the density of water. 3. less than half that of water. 4. more than half the density of water. 5. … not enough information is given. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

63 1. rises. 2. falls. 3. remains in place.
In the presence of air, the small iron ball and large plastic ball balance each other. When air is evacuated from the container, the larger ball 1. rises. 2. falls. 3. remains in place. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

64 1. half. 2. the same. 3. twice. 4. more than twice.
The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be 1. half. 2. the same. 3. twice. 4. more than twice. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

65 Ch09b - Fluids-Revised: 3/6/2010
A pair of identical balloons are inflated with air and suspended on the ends of a stick that is horizontally balanced. When the balloon on the left is punctured, the balance of the stick is 1. upset and the stick rotates clockwise. 2. upset and the stick rotates counter-clockwise. 3. unchanged. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010

66 Ch09b - Fluids-Revised: 3/6/2010
Consider an air-filled balloon weighted so that it is on the verge of sinking—that is, its overall density just equals that of water. Now if you push it beneath the surface, it will 1. sink. 2. return to the surface. 3. stay at the depth to which it is pushed. MFMcGraw Ch09b - Fluids-Revised: 3/6/2010


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