Lecture 21Pressure vs. Depth Incompressible Fluids (liquids)Pressure vs. DepthPressure Measurements: BarometerExercise PressureArchimedes’ Principle: A Eureka MomentArchimedes’ PrincipleSink or Float?Bar TrickExerciseSlide 12Pascal’s PrinciplePascal’s Principle in action: Hydraulics, a force amplifierSlide 20Fluids in MotionTypes of Fluid FlowSlide 24Ideal FluidsExercise ContinuitySlide 27Slide 29Slide 30Physics 207: Lecture 21, Pg 1Lecture 21Goals:Goals:•Chapter 15Chapter 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•AssignmentAssignment HW9, Due Wednesday, Apr. 8th Thursday: Read all of Chapter 16Physics 207: Lecture 21, Pg 2In many fluids the bulk modulus is such that we can treat the density as a constant independent of pressure:An incompressible fluidFor an incompressible fluid, the density is the same everywhere, but the pressure is NOT! p(y) = p0 - y g p0 + d gGauge pressure (subtract p0,pressure 1 atm, i.e. car tires)Pressure vs. DepthIncompressible Fluids (liquids) y1y2Ap1p2F1F2mg0p F2 = F1+ m g = F1+ VgF2 /A = F1/A + Vg/Ap2 = p1 - g yPhysics 207: Lecture 21, Pg 3Pressure vs. DepthFor a uniform fluid in an open container pressure same at a given depth independent of the containerp(y)yFluid level is the same everywhere in a connected container, assuming no surface forcesPhysics 207: Lecture 21, Pg 4Pressure Measurements: BarometerInvented by TorricelliA long closed tube is filled with mercury and inverted in a dish of mercury The closed end is nearly a vacuumMeasures atmospheric pressure as 1 atm = 0.760 m (of Hg)Physics 207: Lecture 21, Pg 5Exercise PressureWhat happens with two fluids?? Consider a U tube containing liquids of density 1 and 2 as shown:At the red arrow the pressure must be the same on either side. 1 x = 2 (d1+ y) Compare the densities of the liquids:(A) 1 < 2 (B) 1 = 2 (C) 1 > 2 12dIyPhysics 207: Lecture 21, Pg 6Archimedes’ Principle: A Eureka MomentSuppose we weigh an object in air (1) and in water (2).How do these weights compare? W2?W1W1 < W2W1 = W2W1 > W2Buoyant force is equal to the weight of the fluid displacedPhysics 207: Lecture 21, Pg 8Archimedes’ PrincipleSuppose we weigh an object in air (1) and in water (2). How do these weights compare? W2?W1W1 < W2W1 = W2W1 > W2 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.Physics 207: Lecture 21, Pg 9Sink 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.F mgByWe can calculate how much of a floating object will be submerged in the liquid: Object is in equilibriummgFBobjectobjectliquidliquidVgVg liquidobjectobjectliquidVVPhysics 207: Lecture 21, Pg 10Bar Trickpiece of rockon top of iceWhat happens to the water level when the ice melts?A. It risesB. It stays the sameC. It dropsExpt. 1 Expt. 2Physics 207: Lecture 21, Pg 11ExerciseV1 = V2 = V3 = V4 = V5m1 < m2 < m3 < m4 < m5What is the final position of each block?Physics 207: Lecture 21, Pg 12ExerciseV1 = V2 = V3 = V4 = V5m1 < m2 < m3 < m4 < m5What is the final position of each block?Not thisBut thisPhysics 207: Lecture 21, Pg 18Pascal’s PrincipleSo far we have discovered (using Newton’s Laws): Pressure depends on depth: p = g yPascal’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 21, Pg 19Pascal’s Principle in action: Hydraulics, a force amplifierConsider the system shown: A downward force F1 is applied to the piston of area A1. This force is transmitted through the liquid to create an upward force F2. Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid.FF12d2d1AA21F2 > F1 with conservation of energyPhysics 207: Lecture 21, Pg 20A1A10A2A10MMdBdAExerciseConsider 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 di in the liquid levels. If A2 = 2A1, how do dA and dB compare? V10 = V1 = V2 dA A1 = dB A2 dA A1 = dB 2A1 dA = dB 2Physics 207: Lecture 21, Pg 22Fluids in MotionTo describe fluid motion, we need something that describes flow: Velocity vThere are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / idealPhysics 207: Lecture 21, Pg 23Types of Fluid FlowLaminar 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 streamlineTurbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speedPhysics 207: Lecture 21, Pg 24Types of Fluid FlowLaminar 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 streamlineTurbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speedPhysics 207: Lecture 21, Pg 25Flow obeys continuity equationVolume flow rate (m3/s) Q = A·v is constant along flow tube.Follows from mass conservation if flow is incompressible.Mass flow rate is just Q (kg/s)A1v1 = A2v2Ideal FluidsStreamlines do not meet or crossVelocity vector is tangent to streamlineVolume of fluid follows a tube of flow bounded by streamlinesStreamline density is proportional to velocityA1A2v1v2Physics 207: Lecture 21, Pg 26 Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the
View Full Document
Unlocking...