Physics 207 Lecture 18


 Carmel Anthony
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1 Physics 07, Lecture 8, Nov. 6 MidTerm Mean 58.4 (64.6) Median 58 St. Dev. 6 (9) High 94 Low 9 Nominal curve: (conservative) A 679 B or A/B 346 C or B/C 933 marginal 98 D Physics 07: Lecture 8, Pg Physics 07, Lecture 8, Nov. 6 Agenda: Chapter 4, luids Pressure, Work Pascal s Principle Archimedes Principle luid flow Assignments: Problem Set 7 due Nov. 4, Tuesday :59 PM Note: Ch. 4:,8,0,30,5a,54 (look at ) Ch. 5:,9,36,4,49 Honors: Ch. 4: 58 or Wednesday, Read Chapter 5 Physics 07: Lecture 8, Pg luids (Chapter 4) At ordinary temperature, matter exists in one of three states Solid  has a shape and forms a surface Liquid  has no shape but forms a surface Gas  has no shape and forms no surface What do we mean by fluids? luids are substances that flow. substances that take the shape of the container Atoms and molecules are free to move. No long range correlation between positions. Physics 07: Lecture 8, Pg 3 Some definitions Elastic properties of solids : Young s modulus: measures the resistance of a solid to a change in its length. L 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 s to changes in their volume. volume elasticity L elasticity of shape (ex. pushing a book) elasticity in length  Physics 07: Lecture 8, Pg 4 luids What parameters do we use to describe fluids? Density ρ = m units : kg/m 3 = 03 g/cm 3 ρ(water) =.000 x 0 3 kg/m 3 =.000 g/cm 3 ρ(ice) = 0.97 x 0 3 kg/m 3 = 0.97 g/cm 3 ρ(air) =.9 kg/m 3 =.9 x 03 g/cm 3 ρ(hg) = 3.6 x0 3 kg/m 3 = 3.6 g/cm 3 luids What parameters do we use to describe fluids? Pressure p = A units : N/m = Pa (Pascal) bar = 0 5 Pa mbar = 0 Pa torr = 33.3 Pa Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. orce (a vector) in a fluid can be expressed in terms of pressure (a scalar) as: n r = panˆ A atm =.03 x0 5 Pa = 03 mbar = 760 Torr = 4.7 lb/ in (=PSI) Physics 07: Lecture 8, Pg 5 Physics 07: Lecture 8, Pg 6 Page
2 Pressure vs. Depth Incompressible luids (s) When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid or an incompressible fluid, the density is the same everywhere, but the pressure is NOT! p p 0 y y A mg p Pressure vs. Depth or a uniform fluid in an open container pressure same at a given depth independent of the container luid level is the same everywhere in a connected container, assuming no surface forces 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. 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. y p(y) Physics 07: Lecture 8, Pg 7 Physics 07: Lecture 8, Pg 8 Pressure Measurements: Barometer Lecture 8, Exercise Pressure Invented by Torricelli A long closed tube is filled with mercury and inverted in a dish of mercury The closed end is nearly a vacuum Measures atmospheric pressure as One atm = m (of Hg) What happens with two fluids?? Consider a U tube containing s of density ρ and ρ as shown: Compare the densities of the s: (A) ρ < ρ (B) ρ = ρ (C) ρ > ρ ρ d I ρ Physics 07: Lecture 8, Pg 9 Physics 07: Lecture 8, Pg 0 Pascal s Principle So far we have discovered (using Newton s Laws): Pressure depends on depth: p = ρ g y 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 and to the walls of the containing vessel. Pascal s Principle explains the working of hydraulic lifts i.e., the application of a small force at one place can result in the creation of a large force in another. Will this hydraulic lever violate conservation of energy? No Pascal s Principle Consider the system shown: A downward force is applied to the piston of area A. This force is transmitted through the to create an upward force. Pascal s Principle says that increased pressure from ( /A ) is transmitted throughout the. d A A > : Is there conservation of energy? d Physics 07: Lecture 8, Pg Physics 07: Lecture 8, Pg Page
3 Lecture 8, Exercise Hydraulics Lecture 8, Exercise 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 d i in the levels. If A = A, compare d A and d B. d A M A A 0 d B M Consider the systems shown on right. If A = A, compare d A and d B. Mg = ρ d A A and Mg = ρ d B A d A A = d B A d A = d B d A M A A 0 d B M (A) d A = (/) d B (B) d A = d B (C) d A = d B (A) d A = (/) d B (B) d A = d B (C) d A = d B If A 0 = A 0, compare d A and d C. A A 0 d C M If A 0 = A 0, compare d A and d C. Mg = ρ d A A and Mg = ρ d c A A A 0 d C M (A) d A = (/) d C (B) d A = d C (C) d A = d C (A) d A = (/) d C (B) d A = d C (C) d A = d C A A 0 A A 0 Physics 07: Lecture 8, Pg 3 Physics 07: Lecture 8, Pg 4 Archimedes Principle Suppose we weigh an in air () and in water (). How do these weights compare? W < W W = W W > W Sink or loat? The buoyant force is equal to the weight of the that is displaced. If the buoyant force is larger than the weight of the, it will float; otherwise it will sink. B mg y Why? Since the pressure at the bottom of the is greater than that at the top of the, the water exerts a net upward force, the buoyant force, on the. W W? We can calculate how much of a floating will be submerged in the : Object is in equilibrium B = mg ρ g = ρ g ρ = ρ Physics 07: Lecture 8, Pg 5 Physics 07: Lecture 8, Pg 6 Lecture 8, Exercise 3 Buoyancy A lead weight is fastened to a large block and the combination floats on water with the water level with the top of the block as shown. If you turn the + upsidedown, What happens? Lecture 8, 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? Cup I Cup II (A) Cup I (B) Cup II (C) the same (D) can t tell (A) It sinks (B) (C) (D) Active igure Physics 07: Lecture 8, Pg 7 Physics 07: Lecture 8, Pg 8 Page 3
4 Lecture 8, Exercise 5 Even More Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (ρ oil < ρ ball < ρ water ) is slowly added to the container until it just covers the ball. Relative to the water level, the ball will: Hint : What is the bouyant force of the part in the oil as compared to the air? water (A) move up (B) move down (C) stay in same place oil luids in Motion Up to now we have described fluids in terms of their static properties: Density ρ Pressure p To describe fluid motion, we need something that can describe flow: elocity v There are different kinds of fluid flow of varying complexity nonsteady / steady compressible / incompressible rotational / irrotational viscous / ideal Physics 07: Lecture 8, Pg 9 Physics 07: Lecture 8, Pg 0 Types of luid low 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 Physics 07: Lecture 8, Pg Types of luid low 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 Physics 07: Lecture 8, Pg Onset of Turbulent low The SeaWifS satellite image of a von Karman vortex around Guadalupe Island, August 0, 999 Ideal luids luid dynamics is very complicated in general (turbulence, vortices, etc.) Consider the simplest case first: the Ideal luid No viscosity  no flow resistance (no internal friction) Incompressible  density constant in space and time Simplest situation: consider ideal fluid moving with steady flow  velocity at each point in the flow is constant in time streamline A A In this case, fluid moves on streamlines v v Physics 07: Lecture 8, Pg 3 Physics 07: Lecture 8, Pg 4 Page 4
5 Ideal luids Streamlines do not meet or cross elocity vector is tangent to streamline olume of fluid follows a tube of flow bounded by streamlines Streamline density is proportional to velocity low obeys continuity equation streamline A A v v Lecture 8 Exercise 6 Continuity A housing contractor saves some money by reducing the size of a pipe from diameter to / diameter at some point in your house. v v / Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the diameter pipe, how fast is the water going in the / pipe? olume flow rate Q = A v is constant along flow tube. (A) v (B) 4 v (C) / v (D) /4 v A v = A v ollows from mass conservation if flow is incompressible. Physics 07: Lecture 8, Pg 5 Physics 07: Lecture 8, Pg 6 Lecture 8 Exercise 6 Continuity A housing contractor saves some money by reducing the size of a pipe from diameter to / diameter at some point in your house. v v / (A) v (B) 4 v (C) / v (D) /4 v or 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 6 times as much kinetic energy. Something must be doing work on the water (the pressure drops at the neck and we recast the work as P = (/A) (A x) = x ) Physics 07: Lecture 8, Pg 7 Conservation of Energy for Ideal luid Recall the standard workenergy relation W = K = K f  K i Apply the principle to a section of flowing fluid with volume and mass m = ρ (here W is work done on fluid) Net work by pressure difference over x ( x = v t) W = x x = ( /A ) (A x ) ( /A ) (A x ) = P P v and = = (incompressible) W = (P P ) and y W = ½ m v ½ m v v p = ½ (ρ ) v ½ (ρ ) v (P P ) = ½ ρ v ½ ρ v y p P + ½ ρ v = P + ½ ρ v = const. Bernoulli Equation P + ½ ρ v + ρ g y = constant Physics 07: Lecture 8, Pg 8 Lecture 8 Exercise 7 Bernoulli s Principle A housing contractor saves some money by reducing the size of a pipe from diameter to / diameter at some point in your house. v v / ) What is the pressure in the / pipe relative to the pipe? (A) smaller (B) same (C) larger Physics 07: Lecture 8, Pg 9 Applications of luid Dynamics Streamline flow around a moving airplane wing Lift is the upward force 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, and the angle between the wing and 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. Physics 07: Lecture 8, Pg 30 Page 5
6 enturi Bernoulli s Eq. Cavitation enturi result In the vicinity of high velocity fluids, the pressure can gets so low that the fluid vaporizes. Physics 07: Lecture 8, Pg 3 Physics 07: Lecture 8, Pg 3 Lecture 8, Recap Agenda: Chapter 4, luids Pressure, Work Pascal s Principle Archimedes Principle luid flow Assignments: Problem Set 7 due Nov. 4, Tuesday :59 PM Note: Ch. 4:,8,0,30,5a,54 (look at ) Ch. 5:,9,36,4,49 Honors: Ch. 4: 58 or Wednesday, Read Chapter 5 Physics 07: Lecture 8, Pg 33 Page 6
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