Fisica Generale - Alan Giambattista, Betty McCarty Richardson Chapter 9: Fluids •Introduction to Fluids •Pressure •Pascal’s Principle •Gravity and Fluid Pressure •Measurement of Pressure •Archimedes’ Principle •Continuity Equation •Bernoulli’s Equation •Viscosity and Viscous Drag •Surface Tension Copyright © 2008 – The McGraw-Hill Companies s.r.l. 1 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.1 Fluids A liquid will flow to take the shape of the container that holds it. A gas will completely fill its container. Fluids are easily deformable by external forces. A liquid is incompressible. Its volume is fixed and is impossible to change. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 2 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.2 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 3 Fisica Generale - Alan Giambattista, Betty McCarty Richardson By Newton’s 3rd Law, there is a force on the wall due to the particle. F Pressure is defined as P . A The units of pressure are N/m2 and are called Pascals (Pa). Note: 1 atmosphere (atm) = 101.3 kPa Copyright © 2008 – The McGraw-Hill Companies s.r.l. 4 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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? 2 1m 4 2 1.0 cm 1.0 10 m 100 cm 2 F 500 N Pav A 1.0 10-4 m 2 1 atm 6 2 1 Pa 5.0 10 N/m 2 5 1 N/m 1.013 10 Pa 49 atm Copyright © 2008 – The McGraw-Hill Companies s.r.l. 5 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.3 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.) Copyright © 2008 – The McGraw-Hill Companies s.r.l. 6 Fisica Generale - Alan Giambattista, Betty McCarty Richardson The applied force is transmitted to the piston of cross-sectional area A2 here. Apply a force F1 here to a piston of crosssectional area A1. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 7 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Mathematically, P at point 1 P at point 2 F1 F 2 A1 A 2 A2 F1 F2 A1 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 8 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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: A2 A1 1 10 100 F2 500 N 5000 N 50,000 N Copyright © 2008 – The McGraw-Hill Companies s.r.l. 9 Fisica Generale - Alan Giambattista, Betty McCarty Richardson The work done pressing the smaller piston (#1) equals the work done by the larger piston (#2). F1d1 F2 d 2 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 10 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example: In the previous example, for the case A2/A1 = 10, it was found that F2/F1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller piston be depressed? Using the result on the previous slide, F2 d1 d 2 10 m F1 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 11 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.4 Gravity’s Effect on Fluid Pressure FBD for the fluid cylinder P1A A cylinder of fluid w P2A Copyright © 2008 – The McGraw-Hill Companies s.r.l. 12 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Apply Newton’s 2nd Law to the fluid cylinder: F P A PA w 0 2 1 P2 A P1 A Ad g 0 P2 P1 gd 0 P2 P1 gd or P2 P1 gd 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 13 Fisica Generale - Alan Giambattista, Betty McCarty Richardson If the top of the fluid column is placed at the surface of the fluid, then P1=Patm if the container is open. P Patm gd Copyright © 2008 – The McGraw-Hill Companies s.r.l. 14 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example (text problem 9.15): 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? P Patm gd P P Patm gd 1000 kg/m 3 9.8 m/s 2 35 m 343 kPa 3.4 atm Copyright © 2008 – The McGraw-Hill Companies s.r.l. 15 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. P Patm gd 95 atm 1 atm gd gd 94 atm 9.5 106 N/m 2 1025 kg/m 9.8 m/s d 9.5 10 3 2 6 N/m 2 d 950 m Copyright © 2008 – The McGraw-Hill Companies s.r.l. 16 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.5 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 17 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 18 Fisica Generale - Alan Giambattista, Betty McCarty Richardson From the figure: At point C Also Pc Patm PB PB' The pressure at point B is the pressure of the gas. PB PB ' PC gd PB PC PB Patm gd Pgauge gd Copyright © 2008 – The McGraw-Hill Companies s.r.l. 19 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 20 Fisica Generale - Alan Giambattista, Betty McCarty Richardson From the figure and PA PB Patm PA gd Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 21 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example (text problem 9.22): An IV is connected to a patient’s vein. The blood pressure in the vein has a gauge pressure of 12 mm of mercury. At least how far above the vein must the IV bag be placed in order for fluid to flow into the vein? Assume that the density of the IV fluid is the same as blood. Blood: Pgauge Hg gh1 h1 12 mm IV: The pressure is equivalent to raising a column of mercury 12 mm tall. Pgauge bloodgh2 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 22 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example continued: At a minimum, the gauge pressures must be equal. When h2 is large enough, fluid will flow from high pressure to low pressure. Pgauge Hg gh1 bloodgh2 Hg gh1 h2 bloodg Hg 13,600 kg/m 3 h1 12 mm 3 1060 kg/m blood 154 mm Copyright © 2008 – The McGraw-Hill Companies s.r.l. 23 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.6 Archimedes’ Principle F1 An FBD for an object floating submerged in a fluid. w F2 The total force on the block due to the fluid is called the buoyant force. FB F2 F1 where F2 F1 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 24 Fisica Generale - Alan Giambattista, Betty McCarty Richardson The magnitude of the buoyant force is: FB F2 F1 P2 A P1 A P2 P1 A From before: P2 P1 gd The result is FB gdA gV Copyright © 2008 – The McGraw-Hill Companies s.r.l. 25 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. FB gV Copyright © 2008 – The McGraw-Hill Companies s.r.l. 26 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example (text problem 9.30): A flat-bottomed barge loaded with coal has a mass of 3.0105 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? F F B FB w FB w0 mw g wVw g mb g wVw mb FBD for the barge w w Ad mb mb 3.0 105 kg 1.5 m d 3 w A 1000 kg/m 20.0 m *10.0 m Copyright © 2008 – The McGraw-Hill Companies s.r.l. 27 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example (text problem 9.34): 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.) FB FBD for the metal w F F B w ma The buoyant force is the weight of the fluid displaced by the object FB waterVg ρ waterV ρ waterVg FB g g g 1 Solve for a: a ρ V m ρ objectVobject object object Copyright © 2008 – The McGraw-Hill Companies s.r.l. 28 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example continued: Since the object is completely submerged V=Vobject. specific gravity water where water = 1000 kg/m3 is the density of water at 4 °C. object 5.0 Given specific gravity water ρ waterV 1 1 ag 1 g 1 g 1 7.8 m/s 2 ρ V S .G. 5.0 object object Copyright © 2008 – The McGraw-Hill Companies s.r.l. 29 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.7 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 v1v2 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 30 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 31 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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 crosssectional area A2. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 32 Fisica Generale - Alan Giambattista, Betty McCarty Richardson m Av t is the mass flow rate (units kg/s) V Av t is the volume flow rate (units m3/s) The continuity equation is 1 A1v1 2 A2v2 If the fluid is incompressible, then 1= 2. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 33 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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 0.2 cm. How fast does the water move through the constriction? A1v1 A2 v2 A1 r12 v2 v1 2 v1 A2 r2 2 1.0 cm 2.0 m/s 50 m/s 0.2 cm Copyright © 2008 – The McGraw-Hill Companies s.r.l. 34 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.8 Bernoulli’s Equation Bernoulli’s equation is a statement of energy conservation. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 35 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 1 2 1 2 P1 gy1 v1 P2 gy2 v2 2 2 Work per unit volume done by the fluid Potential energy per unit volume Kinetic energy per unit volume Points 1 and 2 must be on the same streamline Copyright © 2008 – The McGraw-Hill Companies s.r.l. 36 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example (text problem 9.50): A nozzle is connected to a horizontal hose. The nozzle shoots out water moving at 25.0 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. 1 2 1 2 P1 gy1 v1 P2 gy2 v2 2 2 The hose is horizontal so y1=y2. Also P2 =Patm. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 37 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example continued: Substituting: 1 2 1 2 P1 v1 Patm v2 2 2 1 2 1 2 P1 Patm v2 v1 2 2 v2 = 25m/s and v1 is unknown. Use the continuity equation. d 2 2 2 A2 d2 2 v1 v2 v v2 2 2 A1 d1 d1 2 Since d2<<d1 it is true that v1<<v2. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 38 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example continued: P1 Patm 1 2 1 2 v2 v1 2 2 1 1 2 2 2 v2 v1 v2 2 2 1 2 1000 kg/m 3 25.0 m/s 2 3.1 105 Pa Copyright © 2008 – The McGraw-Hill Companies s.r.l. 39 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.9 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 40 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. V P L 4 r t 8 where is the viscosity Copyright © 2008 – The McGraw-Hill Companies s.r.l. 41 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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 2.0010-3 Pa sec; it is injected into a vein at a gauge pressure of 16.0 mm Hg. (a) What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of 0.150 mL/sec? Solve Poiseuille’s Law for the pressure difference: 3 8L V 8 2.00 10 2 Pa sec 3.00 cm cm P 4 0.15 4 s 1 r t 0.3 10 cm 2830 Pa Copyright © 2008 – The McGraw-Hill Companies s.r.l. 42 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example continued: This pressure difference is between the fluid in the syringe and the fluid in the vein; it is the given gauge pressure. P Ps Pv Ps Pv P 2130 Pa 2830 Pa 4960 Pa Copyright © 2008 – The McGraw-Hill Companies s.r.l. 43 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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 = 0.496 N. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 44 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.10 Viscous Drag The viscous drag force on a sphere is given by Stokes’ law. FD 6rv Copyright © 2008 – The McGraw-Hill Companies s.r.l. 45 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example (text problem 9.64): 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? y FBD for sphere FB FD x w Apply Newton’s Second Law to the sphere F F D FB w ma Copyright © 2008 – The McGraw-Hill Companies s.r.l. 46 Fisica Generale - Alan Giambattista, Betty McCarty Richardson Example continued: When v = vterminal, a = 0 and FD FB w 0 6rvt ml g ms g 0 6rvt lVl g ms g 0 6rvt lVs g ms g 0 Solving for ms g lVs g 2.4 Pa sec 6rvt Copyright © 2008 – The McGraw-Hill Companies s.r.l. 47 Fisica Generale - Alan Giambattista, Betty McCarty Richardson §9.11 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. Copyright © 2008 – The McGraw-Hill Companies s.r.l. 48 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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. 2 F PA r 2 6 r 9 10 N Copyright © 2008 – The McGraw-Hill Companies s.r.l. 49 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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 onesixth the weight of the insect. F 1 6F w or m 5 10 6 kg 6 g Copyright © 2008 – The McGraw-Hill Companies s.r.l. 50 Fisica Generale - Alan Giambattista, Betty McCarty Richardson 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 Copyright © 2008 – The McGraw-Hill Companies s.r.l. 51