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Chapter 9 Fluids

Chapter 9 Fluids. 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. Pressure.

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Chapter 9 Fluids

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  1. Chapter 9Fluids

  2. 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 Ch09e - Fluids-Revised: 7/12/2010

  3. 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. Ch09e - Fluids-Revised: 7/12/2010

  4. 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) = 101.3 kPa Ch09e - Fluids-Revised: 7/12/2010

  5. 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? A 500N person weighs about 113 lbs. Ch09e - Fluids-Revised: 7/12/2010

  6. y x Gravity’s Effect on Fluid Pressure FBD for the fluid cylinder An imaginary cylinder of fluid P1A Imaginary cylinder can be any size w P2A Ch09e - Fluids-Revised: 7/12/2010

  7. Apply Newton’s 2nd Law to the fluid cylinder. Since the fluids isn’t moving the net force is zero. 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. Ch09e - Fluids-Revised: 7/12/2010

  8. If the top of the fluid column is placed at the surface of the fluid, then P1 = Patm if the container is open. You noticed on the previous slide that the areas canceled out. Only the height matters since that is the direction of gravity. Think of the pressure as a force density in N/m2 Ch09e - Fluids-Revised: 7/12/2010

  9. 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? Ch09e - Fluids-Revised: 7/12/2010

  10. 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. Ch09e - Fluids-Revised: 7/12/2010

  11. Ch09e - Fluids-Revised: 7/12/2010

  12. Measuring Pressure A manometer is a U-shaped tube that is partially filled with liquid, usually Mercury (Hg). Both ends of the tube are open to the atmosphere. Ch09e - Fluids-Revised: 7/12/2010

  13. 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. Ch09e - Fluids-Revised: 7/12/2010

  14. From the figure: At point C Also The pressure at point B is the pressure of the gas. Ch09e - Fluids-Revised: 7/12/2010

  15. 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. Ch09e - Fluids-Revised: 7/12/2010

  16. From the figure and Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall. Ch09e - Fluids-Revised: 7/12/2010

  17. The Many Units of Pressure 1 ATM equals 1.013x105 N/m2 14.7 lbs/in2 1.013 bar 76 cm Hg 760 mm Hg 760 Torr 34 ft H2O 29.9 in Hg Ch09e - Fluids-Revised: 7/12/2010

  18. 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.) Ch09e - Fluids-Revised: 7/12/2010

  19. 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. In these problems neglect pressure due to columns of fluid. Ch09e - Fluids-Revised: 7/12/2010

  20. Mathematically, Ch09e - Fluids-Revised: 7/12/2010

  21. F2 1 500 N 10 5000 N 100 50,000 N 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: Ch09e - Fluids-Revised: 7/12/2010

  22. F1 y w F2 x Archimedes’ Principle An FBD for an object floating submerged in a fluid. The total force on the block due to the fluid is called the buoyant force. Ch09e - Fluids-Revised: 7/12/2010

  23. Buoyant force = the weight of the fluid displaced The magnitude of the buoyant force is: From before: The result is Ch09e - Fluids-Revised: 7/12/2010

  24. 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. Ch09e - Fluids-Revised: 7/12/2010

  25. FB y w x 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? Apply Newton’s 2nd Law to the barge: FBD for the barge Ch09e - Fluids-Revised: 7/12/2010

  26. FB y w x 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.) 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: Ch09e - Fluids-Revised: 7/12/2010

  27. Example continued: Since the object is completely submerged V=Vobject. where water = 1000 kg/m3 is the density of water at 4 °C. Given The sign is minus because gravity acts down. BF causes a < g. Ch09e - Fluids-Revised: 7/12/2010

  28. V1 = constant V2 = constant v1v2 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. Ch09e - Fluids-Revised: 7/12/2010

  29. 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. Crossing streamlines would indicate a volume of fluid with two different velocities at the same time. An ideal fluid is incompressible, undergoes laminar flow, and has no viscosity. Ch09e - Fluids-Revised: 7/12/2010

  30. The Continuity Equation—Conservation of Mass Faster Slower 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. Ch09e - Fluids-Revised: 7/12/2010

  31. 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. Ch09e - Fluids-Revised: 7/12/2010

  32. 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? Simple ratios Ch09e - Fluids-Revised: 7/12/2010

  33. Bernoulli’s Equation Bernoulli’s equation is a statement of energy conservation. Ch09e - Fluids-Revised: 7/12/2010

  34. This is the most general equation 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 Ch09e - Fluids-Revised: 7/12/2010

  35. 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. Ch09e - Fluids-Revised: 7/12/2010

  36. Example continued: Substituting: v2 = 25 m/s and v1 is unknown. Use the continuity equation. Since d2<<d1 it is true that v1<<v2. Ch09e - Fluids-Revised: 7/12/2010

  37. Example continued: Ch09e - Fluids-Revised: 7/12/2010

  38. 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. Ch09e - Fluids-Revised: 7/12/2010

  39. 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. 4th power where  is the viscosity Ch09e - Fluids-Revised: 7/12/2010

  40. Example (text problem 9.55): A hypodermic syringe attached to a needle has an internal radius of 0.300 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 0.250 mL/sec? Solve Poiseuille’s Law for the pressure difference: Ch09e - Fluids-Revised: 7/12/2010

  41. Example continued: This pressure difference is between the fluid in the syringe and the fluid in the vein. The pressure in the syringe is Conversion: Ch09e - Fluids-Revised: 7/12/2010

  42. 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 = 0.686 N. Ch09e - Fluids-Revised: 7/12/2010

  43. Viscous Drag The viscous drag force on a sphere is given by Stokes’ law. Where  is the viscosity of the fluid that the sphere is falling through, r is the radius of the sphere, and v is the velocity of the sphere. Ch09e - Fluids-Revised: 7/12/2010

  44. y FB FD x w 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? Apply Newton’s Second Law to the sphere FBD for sphere Drag Buoyant Ch09e - Fluids-Revised: 7/12/2010

  45. Example continued: When v = vterminal , a = 0 and Solving for  Ch09e - Fluids-Revised: 7/12/2010

  46. Surface Tension The surface of a fluid acts like 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. Ch09e - Fluids-Revised: 7/12/2010

  47. 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. Ch09e - Fluids-Revised: 7/12/2010

  48. 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. Ch09e - Fluids-Revised: 7/12/2010

  49. 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 Ch09e - Fluids-Revised: 7/12/2010

  50. Quick Questions

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