Physics 207, Lecture 18, Nov. 6 MidTerm 2 Mean 58.4 (64.6) Median 58 St. Dev. 16

Physics 207, Lecture 18, Nov. 6 MidTerm 2 Mean 58.4 (64.6) Median 58 St. Dev. 16 www.phwiki.com

Physics 207, Lecture 18, Nov. 6 MidTerm 2 Mean 58.4 (64.6) Median 58 St. Dev. 16

Stark, Nick, Features Editor has reference to this Academic Journal, PHwiki organized this Journal Physics 207, Lecture 18, Nov. 6 MidTerm 2 Mean 58.4 (64.6) Median 58 St. Dev. 16 (19) High 94 Low 19 Nominal curve: (conservative) 80-100 A 62-79 B or A/B 34-61 C or B/C 29-33 marginal 19-28 D Physics 207, Lecture 18, Nov. 6 Agenda: Chapter 14, Fluids Pressure, Work Pascal’s Principle Archimedes’ Principle Fluid flow Assignments: Problem Set 7 due Nov. 14, Tuesday 11:59 PM Note: Ch. 14: 2,8,20,30,52a,54 (look at 21) Ch. 15: 11,19,36,41,49 Honors: Ch. 14: 58 For Wednesday, Read Chapter 15 Fluids (Chapter 14) At ordinary temperature, matter exists in one of three states Solid – has a shape in addition to as long as ms a surface Liquid – has no shape but as long as ms a surface Gas – has no shape in addition to as long as ms no surface What do we mean by “fluids” Fluids are “substances that flow” . “substances that take the shape of the container” Atoms in addition to molecules are free to move. No long range correlation between positions.

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Some definitions Young’s modulus: measures the resistance of a solid to a change in its length. Shear modulus: measures the resistance to motion of the planes of a solid sliding past each other. Bulk modulus: measures the resistance of solids or liquids to changes in their volume. Elastic properties of solids : elasticity in length elasticity of shape (ex. pushing a book) volume elasticity Fluids What parameters do we use to describe fluids Density units : kg/m3 = 10-3 g/cm3 r(water) = 1.000 x 103 kg/m3 = 1.000 g/cm3 r(ice) = 0.917 x 103 kg/m3 = 0.917 g/cm3 r(air) = 1.29 kg/m3 = 1.29 x 10-3 g/cm3 r(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3 Fluids What parameters do we use to describe fluids Pressure Any as long as ce exerted by a fluid is perpendicular to a surface of contact, in addition to is proportional to the area of that surface. Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as: units : 1 N/m2 = 1 Pa (Pascal) 1 bar = 105 Pa 1 mbar = 102 Pa 1 torr = 133.3 Pa 1 atm = 1.013 x105 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in2 (=PSI)

Pressure vs. Depth Incompressible Fluids (liquids) When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! Pressure vs. Depth If the pressures were different, fluid would flow in the tube! However, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium. For a uni as long as m fluid in an open container pressure same at a given depth independent of the container Fluid level is the same everywhere in a connected container, assuming no surface as long as ces Why is this so Why does the pressure below the surface depend only on depth if it is in equilibrium Imagine a tube that would connect two regions at the same depth. Pressure Measurements: Barometer Invented by Torricelli A long closed tube is filled with mercury in addition to inverted in a dish of mercury The closed end is nearly a vacuum Measures atmospheric pressure as One 1 atm = 0.760 m (of Hg)

Lecture 18, Exercise 1 Pressure What happens with two fluids Consider a U tube containing liquids of density r1 in addition to r2 as shown: Compare the densities of the liquids: r1 r2 dI Pascal’s Principle So far we have discovered (using Newton’s Laws): Pressure depends on depth: Dp = r g Dy Pascal’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 in addition to to the walls of the containing vessel. Pascal’s Principle explains the working of hydraulic lifts i.e., the application of a small as long as ce at one place can result in the creation of a large as long as ce in another. Will this “hydraulic lever” violate conservation of energy No Pascal’s Principle Consider the system shown: A downward as long as ce F1 is applied to the piston of area A1. This as long as ce is transmitted through the liquid to create an upward as long as ce F2. Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid. F2 > F1 : Is there conservation of energy

Lecture 18, Exercise 2 Hydraulics Consider 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 = 2 A1, compare dA in addition to dB. Lecture 18, Exercise 2 Hydraulics Consider the systems shown on right. If A2 = 2 A1, compare dA in addition to dB. Mg = r dA A1 in addition to Mg = r dB A2 dA A1 = dB A2 dA = 2 dB If A10 = 2 A20, compare dA in addition to dC. Mg = r dA A1 in addition to Mg = r dc A1 Archimedes’ Principle Suppose we weigh an object in air (1) in addition to in water (2). How do these weights compare 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 as long as ce, the buoyant as long as ce, on the object.

Sink or Float The buoyant as long as ce is equal to the weight of the liquid that is displaced. If the buoyant as long as ce is larger than the weight of the object, it will float; otherwise it will sink. We can calculate how much of a floating object will be submerged in the liquid: Object is in equilibrium Lecture 18, Exercise 3 Buoyancy A lead weight is fastened to a large styrofoam block in addition to the combination floats on water with the water level with the top of the styrofoam block as shown. If you turn the styrofoam + Pb upside-down, What happens (A) It sinks (C) (B) (D) Active Figure Lecture 18, Exercise 4 More Buoyancy Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. Which cup weighs more

Lecture 18, Exercise 5 Even More Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (roil < rball < rwater) is slowly added to the container until it just covers the ball. Relative to the water level, the ball will: Hint 1: What is the bouyant as long as ce of the part in the oil as compared to the air Fluids in Motion Up to now we have described fluids in terms of their static properties: Density r Pressure p To describe fluid motion, we need something that can describe flow: Velocity v There are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / ideal Types of Fluid Flow Laminar 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 streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed Types of Fluid Flow Laminar 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 streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed Onset of Turbulent Flow The SeaWifS satellite image of a von Karman vortex around Guadalupe Isl in addition to , August 20, 1999 Ideal Fluids Fluid dynamics is very complicated in general (turbulence, vortices, etc.) Consider the simplest case first: the Ideal Fluid No “viscosity” - no flow resistance (no internal friction) Incompressible - density constant in space in addition to time Simplest situation: consider ideal fluid moving with steady flow - velocity at each point in the flow is constant in time In this case, fluid moves on streamlines Stark, Nick British Weekly, The Features Editor www.phwiki.com

Ideal Fluids Streamlines do not meet or cross Velocity vector is tangent to streamline Volume of fluid follows a tube of flow bounded by streamlines Streamline density is proportional to velocity Flow obeys continuity equation Volume flow rate Q = A·v is constant along flow tube. Follows from mass conservation if flow is incompressible. A1v1 = A2v2 Lecture 18 Exercise 6 Continuity A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe v1 v1/2 Lecture 18 Exercise 6 Continuity A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast. But if the water is moving four times as fast the it has 16 times as much kinetic energy. Something must be doing work on the water (the pressure drops at the neck in addition to we recast the work as P DV = (F/A) (ADx) = F Dx ) v1 v1/2

Conservation of Energy as long as Ideal Fluid Recall the st in addition to ard work-energy relation W = DK = Kf – Ki Apply the principle to a section of flowing fluid with volume DV in addition to mass Dm = r DV (here W is work done on fluid) Net work by pressure difference over Dx (Dx1 = v1 Dt) W = F1 Dx1 – F2 Dx2 = (F1/A1) (A1Dx1) – (F2/A2) (A2 Dx2) = P1 DV1 – P2 DV2 in addition to DV1 = DV2 = DV (incompressible) W = (P1– P2 ) DV in addition to W = ½ Dm v22 – ½ Dm v12 = ½ (rDV) v22 – ½ (rDV) v12 (P1– P2 ) = ½ r v22 – ½ r v12 P1+ ½ r v12 = P2+ ½ r v22 = const. Bernoulli Equation P1+ ½ r v12 + r g y1 = constant Lecture 18 Exercise 7 Bernoulli’s Principle A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. 2) What is the pressure in the 1/2” pipe relative to the 1” pipe v1 v1/2 Applications of Fluid Dynamics Streamline flow around a moving airplane wing Lift is the upward as long as ce on the wing from the air Drag is the resistance The lift depends on the speed of the airplane, the area of the wing, its curvature, in addition to the angle between the wing in addition to the horizontal higher velocity lower pressure lower velocity higher pressure Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid.

Venturi Bernoulli’s Eq. Cavitation Venturi result In the vicinity of high velocity fluids, the pressure can gets so low that the fluid vaporizes. Lecture 18, Recap Agenda: Chapter 14, Fluids Pressure, Work Pascal’s Principle Archimedes’ Principle Fluid flow Assignments: Problem Set 7 due Nov. 14, Tuesday 11:59 PM Note: Ch. 14: 2,8,20,30,52a,54 (look at 21) Ch. 15: 11,19,36,41,49 Honors: Ch. 14: 58 For Wednesday, Read Chapter 15

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